CALCULO DIFERENCIAL AMORCITO

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SOLUCIONARIO DE CALCULO DIFERENCIAL E INTEGRAL DE WILLIAM ANTHONY GRANVILLE Ejercicios resueltos por : LIWINTONG MARQUEZ REYES Problemas “ Pagina 14 “ 1. Dado f (x) = x 3 - 5x 2 - 4x + 20 , demostrar que a. f (1) = 12 f (1) = (1) 3 - 5 (1) 2 - 4 (1) + 20 = 1 - 5 - 4 + 20 = 21 - 9 = 12 f (1) = 12 b. f (5) = 0 f (5) = (5) 3 - 5 (5) 2 - 4 (5) + 20 = 125 - 125 - 20 + 20 = 0 f (5) = 0. c. f (0) = - 2f (3) Primero calculamos f (3) f (3) = (3) 3 –5 (3) 2 - 4 (3) + 20 = 27 - 45 - 12 + 20 = 47 - 57 = f (3) = - 10 Luego, calculamos f (0).
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Transcript of CALCULO DIFERENCIAL AMORCITO

SOLUCIONARIO DE CALCULO DIFERENCIAL E INTEGRAL DE WILLIAM ANTHONY GRANVILLE Ejercicios resueltos por : LIWINTONG MARQUEZ REYES

Problemas Pagina 14 1. Dado f (x) = x3 - 5x2 - 4x + 20 , demostrar que a. f (1) = 12 f (1) = (1)3 - 5 (1)2 - 4 (1) + 20 = 1 - 5 - 4 + 20 = 21 - 9 = 12 f (1) = 12 b. f (5) = 0 f (5) = (5)3 - 5 (5)2 - 4 (5) + 20 = 125 - 125 - 20 + 20 = 0 f (5) = 0. c. f (0) = - 2f (3) Primero calculamos f (3) f (3) = (3)3 5 (3)2 - 4 (3) + 20 = 27 - 45 - 12 + 20 = 47 - 57 = f (3) = - 10 Luego, calculamos f (0). f (0) = (0)3 + 5 (0)2 - 4 (0) + 20 = 0 + 0 - 0 + 20 = 20. f (0) = 20 . Sustituyendo f (3) y f (0) en la funcin original. f (0) = - 2 f (3). 20 = -2 (-10) 20 = + 20.

d. f (7) = 5 f (-1) Primero calculamos f (-1) . f (-1) = (-1)3 -5 (-1)2 - 4 (-1) + 20 = - 1 -5 + 4 + 20 = - 6 + 24 = f (-1) = 18. Luego, calculamos f (7). f (7) = (7)3 - 5 (7)2 - 4 (7) + 20 = 343 - 245 - 28 + 20. f (7) = 363 - 273 = 90. Sustituyendo, f (-1) y f (7) en la funcin original. f (7) = 5. f (-1). 90 = 5 (18). 90 = 90. 2. Si f (x) = 4 - 2x2 + x4, calcular : a. f (0) f (0) = 4 - 2 (0)2 + (0)4 = 4 - 0 + 0 = 4 f (0) = 4. b. f (1) f (1) = 4 - 2 (1)2 + (1)4 = 4 - 2 + 1 = 5 - 2. f (1) = 3. c. f (-1) f (-1) = 4 -2 (-1)2 + (-1)4 = 4 - 2 + 1 = 5 - 2 f (-1) = 3. d. f (2) f (2) = 4 -2 (2)2 + (2)4 = 4 - 8 + 16 = 20 - 8 f (2) = 12. e. f (-2) f (-2) = 4 - 2 (-2)2 + (-2)4 = 4 - 8 + 16 = 20 - 8 = f (-2) = 12.

3. Si f ( )

=

sen 2 + cos . Hallar :

a. f (0) f (0) = sen 2 (0) + cos (0) = sen 0 + cos 0 = 0 + 1 = f (0) = 1. b. f (1/2 ) . f (1/2 ) = sen 2 + cos = sen + cos 900 = 0 + 0 = 0 . 2 2 c. f ( ) f () = sen 2 () + cos = sen 3600 + cos 1800 = 0 + (-1) = -1. f () = -1.4.- Dado f (x) = x3 - 5x2 - 4x + 20 , demostrar que :

f (t + 1) = t3 - 2t2 - 11t + 12. f (t + 1) = (t + 1)3 - 5(t + 1)2 - 4(t + 1) + 20. f (t + 1) = t3 + 3t2 + 3t + 1 - 5(t2 + 2t + 1) - 4t - 4 + 20. f (t + 1) = t3 + 3t2 + 3t + 1 - 5t2 - 10t - 5 - 4t - 4 + 20. Haciendo operaciones: f (t + 1) = t3 - 2t2 - 11t + 12. 5. Dado f (y) = y2 - 2y + 6 , demostrar que : f (y + h) = y2 - 2y + 6 + 2 ( y - 1) h + h2. f (y + h) = (y + h)2 - 2(y + h) + 6. f (y + h) = y2 + 2yh + h2 - 2y - 2h + 6. f (y + h) = y2 - 2y + 6 + 2yh - 2h + h2. f (y + h) = y2 - 2y + 6 + h (2y - 1) + h2. f (y + h) = y2- 2y + 6 + ( 2y - 1) h + h2.

6. Dado f (x) = x3 + 3x , demostrar que f (x + h) - f (x) = 3(x2 + 1) h + 3xh2 + h3. Primero encontramos f (x + h) f (x + h) = (x + h)3 + 3(x + h). f (x + h) = x3 + 3x2h + 3xh2 + h3 + 3x + 3h. Luego : f (x + h) - f (x) f (x + h) - f (x) = x3 + 3x2h + 3xh2 + h3 + 3x + 3h - (x3 + 3x). f (x + h) - f (x) = x3 + 3x2h + 3xh2 + h3 + 3x + 3h - x3 - 3x. Efectuando : f (x + h) - f (x) = 3x2h + 3h + 3xh2 + h3. f (x + h) = 3h (x2 + 1) + 3xh2 + h3. f (x + h) = 3 (x2 + 1) h + 3xh2 + h3. 7. Dado f (x) = 1 , demostrar que : f (x + h) - f (x) = _ 1 . 2 x x + xh Primero encontramos f (x + h) : f (x + h) = 1 . x+h Luego : f (x + h) - f (x) = 1 - 1. x+h x

f (x + h) - f (x) = x - (x + h) (x + h) x f (x + h) - f (x) = x - x - h = _ h . (x + h) x x2 + xh

8. Dado (z) = 4z , demostrar que: (z + 1) - (z) = 3 (z) Primero encontramos (z + 1) Luego: Pero : (z+1) = 4z +1 (z + 1) - (z) = 4z +1 - 4z. (z + 1) - (z) = 4z.4 - 4z. (z + 1) - (z) = 4z (4 - 1) = 4z (3) = 3 (4z). (z) = 4z. (z + 1) - (z) = 3 (z).

9. Si (x) = ar ,demostrar que: (y). (z) = (y + z) (y) = ay (z) = az (y). (z) = ay.az = ay + z Si: (x) = ax (y). (z) = ay + z=

(y + z).

10. Dado (x) = log 1 - x ,demostrar que: (y) + (z) = y + z 1+ x 1+ yz. .

Primero calculamos (y) , sustituyendo en (x): (y) = log 1 - y 1+ y Luego calculamos (z) , sustituyendo en (x) : (z) = log 1 - z 1+z Ahora: (y) + (z) = log 1 - y + log 1 - z 1+y 1+z (y) + (z) = log 1 - y - z + yz= =

.

log (1 - y)(1 - z) (1 + y)(1 + z)

.

(1 + yz) - (y + z) .

1+ y + z + yz

(1+ yz) + (y + z)

Ahora calculamos (y + z) , sustituyendo en : (x) = log 1 - x . (1 + yz) 1+x1-(y + z) (y + z) = log 1 + yz (1 + yz) 1+y+z 1+ yz 1 + yz - (y + z) (1 + yz) = log 1 + yz + y + z (1 + yz)=

=

log (1 + yz) - (y + z) (1 + yz) + (y + z)

(y) + (z) = y + z 1 + yz 11.

log (1 + yz) - (y + z) (1 + yz) + (y + z)

Dado : f (x) = sen x , demostrar que

f (x + 2h) - f (x) = 2 cos (x + h). (sen h) Primero encontramos f (x + 2h) sen (x + 2h) = sen x. cos 2h + cos x. sen 2h. Por Trigonomtria : cos 2x = cos2 x - sen2 x = 1 - 2sen2 x. sen 2x = 2sen x.cos x. sen (x + y) = sen x. cos y + cos x. sen y. Sustituyendo en : sen (x + 2h) sen x (cos2 h - sen2 h) + cos x (2 sen h. cos h) sen x (1 - 2 sen2h) + cos x (2 sen h. cos h) sen x (1 - 2 sen2h) + 2 cos x . sen h. cos h. Luego : f (x) = sen x f (x + 2h) = sen x (1 - 2 sen2h) + 2cos x. sen h. cos h f(x + 2h) - f(x) = sen x(1 - 2 sen2h) + 2cos x . sen h.cos h -sen h Haciendo operaciones , simplificando y ordenando: sen x - 2 sen x. sen2h + 2 cos x. sen h. cos h - sen x 2 cos x. sen h. cos h - 2sen x. sen2h Factorando : 2 sen h (cos x. cos h - sen x. sen h) Pero : segn formula , cos x. cos y - sen x. sen y = cos (x+y) Sustituyendo en : 2 sen h (cos x. cos h - sen x. sen h) 2 sen h [(cos (x + h)] = 2 sen h. cos (x + h) = 2 cos (x + h). sen h.

f (x+2h) - f (x) = 2 cos (x+h). sen h .

Problemas Paginas 21 22 Demostrar cada una de las siguientes igualdades: 2. lim 4x + 5 = 2 x 2x + 3Dividiendo nmerador y denominador por x y luego sustituyendo por .

limx

4x + 5 x x 2x + 3 x x

=

4+ 5 x 2+ 3 x.

=

4+ 5 2+ 3

. =

4 + 0 = 4 = 2. 2+0 2

3. lim 4t2 + 3t + 2 = - 1 3 t0 t + 2t - 6 3

Se sustituye t 0 en el numerador y denominador. lim 4 (0)2 + 3(0) + 2 = 0 + 0 + 2 = 2 3 t0 (0) + 2(0) - 6 0 + 0 - 6 -6 4. lim x2h + 3xh2 + h3 h0 2xh + 5h2= =

-1. 3

x. 2=

lim h (x2 + 3xh + h2 ) h0 h (2x + 5h)

x2 + 3xh + h2 2x + 5h

Se sustituye h 0 tanto en el numerador como en el denominador. lim x2 + 3x(0) + (0)2 = x2 + 0 + 0 h0 2x + 5(0) 2x + 0=

x2 2x

=

x .x 2x

=

x . 2

5. lim 6x3 - 5x2 + 3 3 x 2x + 4x - 7

=

3

Primero dividimos, tanto en el numerador como en el denominador por x3.

6x3 - 5x2 + 3 lim x3 x3 x3 3 x 2x + 4x - 7 3 x x3 x3

=

6- 5+ 3 x x3 2+ 4 - 7 x2 x3

.

=

Luego sustituyendo x y teniendo presente que todo nmero para = 0 .

6- 5 + 3 3 2+ 4 - 7 2 3

. =

6-0+0 2+0-0 1

=

6 =3. 2

6. lim (2z + 3k)3 - 4k2z 2 k 0 2z ( 2z - k )

=

lim (2z)3 + 3(2z)2(3k) + 3(2z)(3k)2 + (3k)3 - 4k2z . k0 2z [(2z)2 - 2zk + (k)2] lim 8z3 + 36z2k + 54zk2 + 27k3 - 4k2z . k0 2z (4z2 - 2zk + k2) Sustituyendo k 0 lim 8z3 + 36z2 (0) + 54z (0)2 + 27 (0)3 - 4(0)2z . k0 2z [4z2 - 2z(0) + (0)2] lim 8z3 + 0 + 0 + 0 - 0 2 k0 2z (4z - 0 + 0)=

8z3 8z3

=

1.

7. limx

ax4 + bx2 + c dx5 + ex3 + fx

=

0

Dividiendo numerador y denominador para x4 . ax4 + bx2 + c lim x4 x4 x4 5 3 x dx + ex + fx x4 x4 x4 limx

=

a+ b + c x2 x4 . dx + e + f x x3

.

x en la operacin..

a+ b + 2 d. + e + a =0

c 4 f 3

.

= a + 0+0 +0+0

.

limx

8. limx

ax4 + bx2 + c dx + ex2 + fx + g3

=

.

Dividiendo numerador y denominador para x4. ax4 + bx2 + c lim x4 x4 x4 3 2 x dx + ex + fx + g 4 4 x x x4 x4 a+ b + c x2 x4 d + e + f + g x x2 x3 x4.

=

.

.

Sustituyendo x en la operacin.limx

a + b + c 2 4 a+0+0 = d+ e + f + g 0+0+0+0 2 3 4

.

=

a 0

=

9. lim s4 - a4 2 2 sa s - a

=

2a2 s2 + a2.

lim (s2 + a2) (s2 - a2) sa ( s2 - a2 )

=

Sustituyendo s a en la operacin. lim a2 + a2 = 2a2sa

10. lim x2 + x - 6 x2 x2 - 4

=

5 4

.

lim (x + 3) (x - 2) = (x + 3) . Sustituyendo x2 : x2 (x + 2) (x - 2) (x +2) lim (2 + 3) = 5 . x2 (2 + 2) 4 11. lim 4y2 - 3 = 0 y 2y3 + 3y2 Dividimos para y3. 4 y2 - 3 4 - 3. 3 3 lim y y y y3 = 3 2 y 2 y + 3 y 2+ 3 3 3 y y y Sustituyendo y en la operacin : 4 - 3 lim y 2 + 3.

.

=

0-0 2+0

=

0 =0 2

12. lim 3h + 2xh2 + x2h3 = - 1 . 3 3 h 4 - 3xh - 2x h 2x Dividiendo todo para h3. 3h + 2xh2 + x2h3 lim h3 h3 h3 = 3 3 h 4 - 3xh - 2x h h3 h3 h3 3 + 2x + x2 h2 h 4 - 3x - 2x3 h3 h2

.

Sustituyendo h en la operacin :

limh

3 + 2x + x2 2 4 - 3x - 2x3 3 2

=

3 + 2x + x2 4 - 3x - 2x3 =

=

0 + 0 + x2 0 - 0 - 2x3

=

x2 -2x3

=

- 1 . 2x

13. lim aoxn + a1xn-1 + + a n n n-1 x box + b1x + + bn

ao . bo

lim aoxn + a1xn.x-1 + + an . Dividiendo todo para el mayor exponente xn x boxn + b1xn.x-1 + + b

aoxn + a1xn.x-1 + + an lim xn xn xn n n -1 x box + b1.x .x + + bn n n x x xnSustituyendo en x.

=

ao + a1.x-1 + + an . xn . bo + b1.x-1 + + bn xn

limx

ao + a1 + + an bo + b1 + + bn

=

limx

ao + 0 + + 0 bo + 0 + + 0

=

ao bo

.

14. lim aoxn + a1xn-1 + + an n n-1 x0 box + b1x + + bnSustituyendo x0 en x

=

an bn

lim ao ( 0 )n + a1 ( 0 )n-1 + + a n x0 b0 ( 0 )n + b1 ( 0 )n-1 + + b n lim 0 + 0 + + an = an . x 0 + 0 + + bn bn

=

ao (0) + a1 (0) + + an . bo (0) + b1 (0) + + bn

15. lim (x + h)n - xn h0 h

=

nxn-1

Desarrollando el Binomio de Newton.

lim xn + nxn-1h + n(n-1).xn-2.h2 + n(n-1)(n-2).xn-3.h3 + + hn - xn. h0 1x2 1x2x3 lim nxn-1.h + n(n-1).xn-2.h2 + n(n-1)(n-2).xn-3.h3 + + hn h0 2 6Dividiendo todo para h .

lim nxn-1. h + n(n-1).xn-2. h 2 + n(n-1)(n-2).xn-3. h 3 + + h n h0 h 2h 6h h lim nxn-1 + n(n-1).xn-2.h + n(n-1)(n-2).xn-3.h2 + + hn-1 h0 2 6Sustituyendo h0 en la operacin.

..

lim nxn-1 + n(n-1).xn-2( 0 ) + n(n-1)(n-2).xn-3( 0 )2 + + ( 0 )n-1 h0 2 6 lim nxn-1 + 0 + 0 + + 0 = nxn-1h0

16. lim x + h - x = 1 . h0 h 2 xRacionalizando el numerador:

lim (x + h - x ) (x + h + x ) h0 h (x + h + x) lim (x + h )2 - (x )2 = x + h - x h0 h(x + h + x) h(x+h + x) limh0

.

h 1 = h (x + h + x) (x + h + x )

.

Sustituyendo h0 en la operacin.

limh0

1 (x + 0 + x )

=

1 = (x + x )

1 . 2x

17. Dado f (x) = x2 , demostrar que : lim f (x+h) - f (x) h0 hSi f (x) = x2=

2x

f (x+h) = (x+h)2 lim (x + h)2 - x2 = x2 + 2xh + h2 - x2 h0 h hSustituyendo h 0 en la operacin:=

2xh + h2 h

=

h (2x + h) h

=.

2x + h

lim 2x + h = 2x + 0 = 2xh0

18. Dado f (x) = ax2 + bx + c , demostrar que:lim f (x + h) - f (x) h0 h=

2ax + b.

f ( x ) = ax2 + bx + c. f (x + h) = a (x + h)2 + b (x + h) + c. f (x + h) = a (x2 + 2xh + h2 ) + bx + bh + c. f (x + h) = ax2 + 2axh + ah2 + bx + bh + c.

Reemplazando en la funcin:lim f (x + h) - f (x) h0 h=

ax2 + 2axh + ah2 + bx + bh + c - (ax2 + bx + c). h

lim ax2 + 2axh + ah2 + bx + bh + c - ax2 - bx - c = 2axh + ah2 + bh . h0 h h lim h (2ax + ah + b ) = 2ax + ah + b ; h0 h lim 2ax + a ( 0 ) + b = 2ax + bh0.

19. Dado f (x) = 1 ,demostrar que : x lim f (x + h) - f (x) h0 h f(x) = 1 . x f(x+h) = 1 . x+h=

- 1 . x2

1 - 1 lim x + h x h0 h limh0

=

x - (x + h) x (x + h) h 1

=

x-x-h x (x + h) h 1

=

-h . x (x + h) h 1

=.

-1 x (x + h).

.

Sustituyendo h 0 en la operacin final: lim 1 1 = - 1 . = h0 x (x + 0) x.x x2 20. Si f (x) = x3 , hallar lim f (x + h) - f (x) = 3x2 h0 h f (x) = x3. f (x + h) = (x + h)3 = x3 + 3x2h + 3xh2 + h3. Reemplazando estos valores: lim f (x+h)3 - f (x) h0 h=

x3 + 3x2h + 3xh2 + h3 - x3 = 3x2h + 3xh2 + h3 . h h

lim h (3x2 + 3xh + h2 ) = 3x2 + 3xh + h2 h0 h.

Sustituyendo h0 en la operacin. lim 3x2 + 3x ( 0 ) + ( 0 )2 = 3x2 + 0 + 0 = 3x2h0

Problemas Pgina 32 Calcular la derivada de cada una de las siguientes funciones usando la regla general. 1. y = 2 - 3xSe sustituye en la funcin "x" por "x + x" y se calcula el nuevo valor de la funcin y + y . y + y = 2 - 3 (x + X) . Se resta el valor dado de la funcin del nuevo valor y se obtiene y. Y + y = 2 - 3 (x + X) Y + y = 2 3X -3X=

y + y - y y = - 3x.

2 - 3x - 3x 2 + 3X

.

Se divide y para x.

y x

=

- 3x x

Se calcula el lmite de este cociente cuando x 0 . El lmite as hallado es la derivada buscada.

y

=

X lim x0

- 3x . x.

dy = - 3 dx

Y = -3

2.

y = mx + b.y + y = m (x + x) + b. Y + y = mX + mx + b y + y - y = mx + mx + b mX - b

lim x0

y = mX y = m x x x y = m x dy = m . dx

Y=m

3.

y = ax2 y + y = a ( X + x)2. y + y = a ( X2 +2X X + x2 ) y + y = aX2 +2aX X + ax2 y + y - y = aX2 +2aX X + ax2 - aX2 y = 2aX X + ax2 y = 2ax. x + a.x2 x x x y = 2ax + a.x . x.

.

lim x0

dy = 2ax + a (0) dx dy = 2ax . dx Y = 2 ax

4.

s = 2t - t2. s + s = 2(t + t) - (t + t)2. s + s = 2t + 2t - (t2 +2t t + t 2 ) s + s = 2t + 2t - t2 -2t t - t 2 s + s - s = 2t + 2t - t2 -2t t - t 2 2t +t2 s = 2. t - 2t. t - t 2 s = t (2 - 2t - t) t t s = 2 - 2t - t.

lim x0

t

ds dt

=

2 - 2t - 0

Y = 2 2t

5.

y = cx3 y + y = c ( x + x)3. y + y = c ( x3 + 3x2. x + 3x.x2 + x3 ) y + y = c x3 + 3cx2. x + 3cx.x2 + c.x3 y + y - y = c x3 + 3cx2. x + 3cx.x2 + c.x3 - c x3 y = 3cx2.x + 3cx.x2 + c x3 y = 3cx2. x + 3cx. x2 + cx3 x x x x .y = 3cx2 + 3cx.x + cx2. x dy = 3cx2 + 3cx( 0 ) + c ( 0 )2 dx

lim x0

Y = 3cx2

6.

y = 3x - x3.y + y = 3 (x + x) - (x + x)3. y + y = 3 x + 3x - (x3 +3X2 x + 3X x2 + x3 ) y + y = 3 x + 3x - x3 -3X2 x - 3X x2 - x3 y + y - y = 3 x + 3x - x3 -3X2 x - 3X x2 - x3 3X + X3. y = 3x - 3X2 x - 3X x2 - x3 y = 3.x - 3x2.x - 3x.(x)2 - (x)3 . x x x x x 2 2 y = 3 - 3x - 3x (0) - (x) x

lim x0

dy = 3 - 3x2 . dx 7. u = 4v2 + 2v3.

Y = 3 3x2

u + u = 4 (v + v)2 + 2 (v + v)3.

u + u u = 4 v2 + 8vv +4v2 + 2v3 + 6v2v + 6vv2 + 2v3 4v2 2v3

u + u = 4 ( v2 + 2vv + v2 ) + 2 ( v3 + 3v2v + 3vv2 + v3 ) u + u = 4 v2 + 8vv +4v2 + 2v3 + 6v2v + 6vv2 + 2v3

u = 8vv +4v2 + 6v2v + 6vv2 + 2v3 u = 8v. v + 4. v2 + 6v2. v + 6v. v2 + 2. v3. v v v v v v

lim v0

u = 8v + 4. v + 6v2 + 6v. v + 2. v2 v u = 8v + 4(0) + 6v2 + 6v(0) + 2(0 )2 v du = 8v + 0 + 6v2 + 0 + 0 dv du = 8v + 6v2 . U = 8v + 6v2

dv 8. y = x4. y + y = (x + x)4. y + y = x4 + 4X3x + 6X2 x2 + 4Xx3 + x4 y + y - y = x4 + 4X3x + 6X2 x2 + 4Xx3 + x4 - x4 y = 4x3. x + 6x2.x2 + 4x.x3 + x4.y = 4x3. x + 6x2(x)2 + 4x(x)3 + (x)4 x x x x x.

lim x0

y = 4x3 + 6x2. x + 4x (x)2 + (x)3 x y = 4x3 + 6x2(0) + 4x(0)2 + ( 0 )3 x dy = 4x3 . dx Y = 4x3

9.

e=

2 . +1=

e + e

2 . ( + ) + 1=

e + e - e

2 ( + ) + 1

2 . (+1)

e = 2 ( + 1) - 2[( + ) + 1] . [( + ) + 1] ( + 1) e e=

=

2 + 2 - 2 - 2. - 2 [( + ) + 1] ( + 1) - 2.

=

=

2. . [( + ) + 1]( + 1) -2 =

e =

[( + ) + 1]( + 1)() -2 . [( + 0) + 1]( + 1) -2 = ( + 1) ( + 1)=

[( + ) + 1]( + 1)

lim x0

de d de d Y

=

-2 . ( + 1)2 - 2 . ( + 1)2

=

10.

y=

3 . x2 + 2 y + y = 3 . 2 (x + x) + 2 y + y - y = 3 - 3 . 2 (x + x) + 2 x2 + 2 y = 3 (x2 + 2) - 3 [(x + x)2 + 2] [(x + x)2 + 2] (x2 + 2) y = 3x2 + 6 -3 [x2 + 2x. x +(x)2 +2] [(x + x)2 + 2] (x2 + 2) y = 3x2 + 6 - 3x2 - 6x.x - 3(x)2 - 6 [(x + x)2 + 2] (x2 + 2) y = x (- 6x -3. x) x [(x + x)2 + 2] (x2 + 2) x= =

- 6x. x - 3(x)2 . [(x+x)2 + 2] (x2 + 2)=

- 6x - 3. x [(x + x)2 + 2] (x2 + 2)

lim x0

y = - 6x - 3 (0) . x [(x + 0)2 + 2] (x2 + 2) dy = - 6x - 0 - 6x . = 2 2 2 dx (x + 2) (x + 2) (x + 2)2

11.

s=t+4 t s + s=

(t + t) + 4 t + t

s + s - s = t + t + 4 - t + 4 t + t t s = t (t + t + 4) - (t + 4) (t + t) (t + t) t=

s = t2 + t.t + 4t -(t2 + 4t + t. t + 4. t) = (t + t) t s = t2 + t. t + 4t - t2 - 4t - t. t - 4. t (t + t) t s t= =

- 4. t . (t + t) t

- 4 ( t ) . (t + t) t ( t )=

st

-4 . (t + 0)t

lim t0

ds = - 4 dt t.t

=

-4 . t2

S = -4 / t2 12. y= 1 . 1 - 2x=

y + y

1 . 1 - 2(x + x) 1 1 . 1 - 2 (x + x) 1 - 2x=

y + y - y =

y = (1 - 2x) - [1 -2(x + x)] [1 - 2(x+x)](1 - 2x) y = 1 - 2x - 1 + 2x + 2x [1 - 2(x+x)](1 - 2x)

1 - 2x -(1 - 2x - 2x) [1 - 2(x+x)](1 - 2x)

=

=

2x . [1 - 2 (x + x)](1 - 2x)

y = 2 x 2 . = x x [1 - 2(x + x)](1 - 2x) [1 - 2(x + x)](1 - 2x) yx=

2 . [1 - 2 (x + 0)](1 - 2x) 2 . 2 (1 - 2x)

lim x0 dy = 2 dx (1 - 2x) (1 - 2x)=

dy = dx 13. e=

2 (1 - 2x)2 +2.

e + e =

+ ( + ) + 2

.

e + e - e =

+ _ ( + ) + 2 +2

.

- 2

e =( + 2) ( + ) - [( + ) + 2] = 2 + 2 + . + 2 - 2 - . .

[( + ) + 2]( + 2)

[( + ) + 2] ( + 2)

e =

2 [( + ) + 2] ( + 2)

.

Dividiendo a ambos miembros para y simplificando : e = 2 2. = [( + ) + 2] ( + 2). [( + ) + 2] ( + 2). e =.

.

2 = [( + 0) + 2] ( + 2)0

2 = ( + 2) ( + 2)

2 . 2 ( + 2)

lim

de = 2 d ( + 2) ( + 2) 14. s = At + B Ct + D

=

2 . 2 ( + 2)

s + s = A(t + t) + B C(t + t) + D

.

s + s - s = A(t + t) + B - At + B [C(t + t) + D] Ct + D s = [A(t + t) + B] (Ct + D) - [C(t + t) + D] (At + B) [C(t + t) + D] (Ct + D) s = [A.t + A.t + B](C.t + D) - [C.t + C.t + D](A.t + B) [C(t + t) + D] (Ct + D) s = ACt2 + ADt + AC . t . t + ADt + BCt + BD [C(t + t) + D] (Ct + D) - ACt2 - BCt - ACtt - BCt - Adt - BD . [C(t + t) + D] (Ct + D)s = s t s A.D. t - B.C. t [C(t + t) + D] (Ct + D) t (A.D - B.C) [C(t + t) + D] (Ct + D)= =.

t (A.D - B.C) [C(t + t) + D] (Ct + D)

.

=

(A.D - B.C) . (t) [C(t + t) + D] (Ct + D).

=

t lim t0

(A.D - B.C) (A.D - B.C) = [C(t + 0) + D] (Ct + D) (Ct + D)(Ct + D)

ds = (A.D - B.C) dt (Ct + D)(Ct + D) 15. y = x3 + 1 x

=

(A.D - B.C) . (Ct + D)2

.

y + y = (x + x)3 + 1 . (x + x) y + y - y = (x + x)3 + 1 - x3 + 1 (x + x) x y = [(x + x)3 + 1] x - (x3 + 1) (x + x) (x + x) xy = {[x3 + 3x2x + 3x(x)2 + (x)3] + 1}(x) - x4 - x3(x) - x - x (x + x) x

y = x4 + 3x3x + 3x2(x)2 + (x)3(x) + x - x4 - x3( x) - x -x (x + x) x y = 2x3. x + 3x2(x)2 + (x)3(x) - 3x2(x)2 + (x)3(x) - x (x + x) x y = x [2x3 + 3x2 x + (x)2(x) - 3x2(x) + (x)2(x) - 1] (x + x) x Dividiendo a ambos miembros para x, tenemos : y = x [2x3 + 3x2 x + (x)2(x) - 3x2(x) + (x)2(x) - 1] x (x + x) (x) (x) y = x [2x3 + 3x2. x + (x)2.x - 3x2. x + (x)2.x - 1] x (x + x)(x)( x) y = [2x3 + 3x2( 0 ) + (0)2(x) - 3x2(0) + (0)2(x) - 1] x (x+0)xlim x0

y = 2x3 + 0 + 0 - 0 + 0 - 1 = 2x3 - 1 = 2x3 - 1 x.x x2 x2 x2

dy = 2x - 1 dx x2 16. y= 1 x + a22 .

y + y =

1 . 2 2 (x + x) + a 1 _ 1 2 2 2 (x + x) + a x + a2=

y + y - y =

.

y = 1 (x2 + a2) - 1 [(x + x)2 + a2] [(x + x)2 + a2] (x2 + a2)

x2 + a2 -[x2 + 2x.x + (x)2 + a2] [(x+x)2 + a2] (x2 + a2)=

y = x2 + a2 -x2 - 2x. x -(x)2 - a2 [(x + x)2 + a2](x2 + a2) y = - x (2x + x) . 2 2 2 2 [(x + x) + a ] (x + a )

- 2x. x -(x)2 . [(x + x)2 + a2] (x2 + a2)

Dividiendo a ambos miembros para x, tenemos :

y = - x (2x + x) x [(x + x)2 + a2](x2 + a2). x yx lim x0=

=

- (2x + x) . 2 2 2 2 [(x + x) + a ] (x + a )

- (2x + 0) [(x + 0)2 + a2] (x2 + a2)

.

dy = - 2x dx (x2 + a2) (x2 + a2) dy = dx - 2x (x + a2 )22

=

- 2x (x2 + a2 )2

.

17.

y=

x . x2 + 1 x + x . (x + x)2 + 1 x + x - x . 2 [(x + x) + 1] (x2 + 1)

y + y =

y + y - y =

y = (x + x) (x2 + 1) - x [(x + x)2 + 1] . [(x + x)2 + 1] (x2 + 1) y = x3 + x + x. x2 + x - x [x2 + 2x. x + (x)2 + 1] [(x + x)2 + 1] (x2 + 1) y = x 3 + x + x. x2 + x - x 3 - 2. x. x2 - x. (x)2 - x . [(x + x)2 + 1] (x2 + 1) y = - x.x2 - x.(x)2 + x [(x + x)2 + 1] (x2 + 1)=

- x (x2 + x . x - 1 ) [(x + x)2 + 1](x2 + 1)

.

y = - x (x2 + x. x - 1) = - (x2 + x. x - 1) x [(x + x)2 + 1](x2 + 1) . x [(x + x)2 + 1](x2 + 1) yx=

.

- [x2 + x(0) - 1] [(x + 0)2 + 1] (x2 + 1)

.

lim x0 dy dx -(x2 - 1) (x + 1) (x2 + 1)2

=

=

1 - x2 (x2 + 1)2

18.

y = x2 . 4 - x2 y + y = (x + x)2 . 4 - (x + x)2 y + y - y = (x + x)2 - x2 . [4 - (x + x)2] (4 - x2)

y = (x + x)2 (4 - x2) - [4 - (x + x)2] x2 . [4 - (x + x)2] (4 - x2) y = [x2 + 2x. x + (x)2](4 - x2) - [4-(x2 +2x. x + (x)2]( x2 ) [4 - (x + x)2] (4 - x2)y = 4x2 + 8x. x + 4( x)2 - x4 - 2x3. x - x2.( x)2 -[4 -x2-2x. x-( x)2](x2 ) [4 - (x + x)2] (4 - x2) y = 4x2 + 8x. x + 4( x)2 x 4- 2x3. x - x2.( x)2 - 4x2 + x 4+ 2x3. x + x2. ( x)2 [4 - (x + x)2](4 - x2)

y =

8x.x + 4. (x)2 [4 - (x + x)2](4 - x2)

=

x (8x + 4. x) . [4 - (x + x)2](4 - x2)

Dividiendo, para x , tenemos :

y = x (8x + 4. x) y [4 - (x + x)2](4 - x2) . xyx lim x0=

. .

8x + 4. x [4 - (x + x)2](4 - x2)

=

8x + 4( 0 ) [4 - (x + 0)2](4 - x2)

=

dy = 8x + 0 dx [4 - x2] (4 - x2 ) 19. y = 3x2 - 4x - 5.

=

8x (4 - x2 )2

y + y = 3 (x + x)2 - 4 (x + x) - 5y + y - y = 3 (x + x)2 - 4 (x + x) - 5 - (3x2 - 4x -5)

y = 3 [x2 + 2x. x + (x)2] - 4 (x + x) - 5 - (3x2 - 4x -5)y = 3x2 + 6x. x + 3.(x)2 - 4x - 4. x - 5 - 3x2 + 4x + 5.

y = 6x. x + 3 (x)2 - 4.(x) = (x) [6x + 3 (x) - 4]Dividiendo para x :

y

=

x lim x0

(x)[6x + 3 (x) - 4] = x [6x + 3 (x) - 4] = 6x + 3(0) - 4 x x

dy = 6x - 4 = 2(3x - 2) dx 20. s = at2 + bt + c. s + s = a (t + t)2 + b (t + t) + c . s + s - s s= =

a (t + t)2 + b (t + t) + c - (at2 + bt + c) .

a [t2 + 2t. t + (t)2] + bt + b.t + c - at2 - bt - c .

s = at2 + 2at. t + a.( t)2 + bt + b. t + c - at2 - bt - c .

s = 2at. t + a.( t)2 + b. t

Dividiendo para t , factorizando y simplificando :

s = t (2at + a. t + b) = t (2at + a. t + b) t t tt lim t0

s

=

2at + a( 0 ) + b = 2at + 0 + b .

ds = 2at + b . dt

21.

u = 2v3 - 3v2u + u = 2 (v + v)3 - 3 (v + v)2 u + u - u=

2(v + v)3 - 3 (v + v)2 - (2v3 - 3v2)

u = 2[v3 + 3v2. v + 3v.( v)2 + (v3)] - 3[v2 + 2v. v + (v)2] - 2v3 + 3v2 u = 2v3 + 6v2. v + 6v (v)2 + 2 (v)3 - 3v2 - 6v. v - 3(v)2 - 2v3 + 3v2 .

u = 6v2. v + 6v (v)2 + 2 (v)3 - 6v. v - 3(v)2 Factorizando y dividiendo para v :

u = v [6v2 + 6v. v + 2. (v)2 - 6v - 3. v] v v .v lim v0

u

=

6v2 + 6v. v + 2 (v)2 - 6v - 3. v

u = 6v2 + 6v (0) + 2 (0)2 - 6v - 3 (0) . v

du = 6v2 - 6v dv 22. y = ax3 + bx2 + cx + d .y + y = a (x + x)3 + b (x + x)2 + c (x + x) + d .y + y - y = [a (x + x)3 + b (x + x)2 + c(x + x) + d] - (ax3 + bx2 + cx + d) .

y + y - y = a[x3 + 3x2x + 3x(x)2 + (x)3] + b[x2 + 2x. x + (x)2 + cx + c. x + d - (ax3 + bx2 + cx + d)

y = ax3 + 3ax2.x + 3ax.(x2) + a.(x)3 + bx2 + 2bx.x + b(x)2 + cx + c. x + d - ax3 - bx2 - cx - d . y = 3ax2.x + 3ax.(x2) + a.(x)3 + 2bx.x + b(x)2 + c. x

Factorizando y dividiendo para x :

y = x (3ax2 + 3ax.x + a.(x)2 + 2bx + b.x + c ) x xx lim v 0

y

=

3ax2 + 3ax ( 0 ) + a.( 0 )2 + 2bx + b.( 0 ).x + c

dy = 3ax2 + 0 + 0 + 2bx + 0 + c dx 23. e = (a - b )2 e + e = [a - b ( + )]2 e + e - e = [a - b ( + )]2 - (a - b)2

e e

=

(a - b - b.)2 - (a - b)2 [a + (-b) + (- b.)]2 - (a - b)2

=

e = a2 + (-b)2 + (- b.)2 + 2a.(-b) +2a.(- b.) +2.(-b).(- b. ) - [a2-2a.b + (b)2]

e = a2 + (b)2 +(b.)2- 2a(b) -2a(b.) +2(b)(b.) - a2 + 2a.b - (b)2

e e

=

(b. )2 -2a(b. ) + 2(b)(b. ) b2()2 - 2a(b. ) + 2(b)(b.)

=

Factorando y dividiendo para . e =

.

{b2.( ) - 2a.b + 2b2.} =

.

e lim

b2.() - 2a.b + 2b2. = b2.(0) - 2a.b + 2b2.

0

de = 0 - 2ab + 2b2. = 2b2. - 2ab = 2b (b. - a ) d 24. y = (2 - x) (1 - 2x) . y + y = [2 - (x + x)] [1 - 2 (x + x)] y + y - y = [2 - (x + x)] [1 - 2 (x + x)] - (2 - x) (1 - 2x) y = (2 - x - x) (1 - 2x - 2.x) - (2 - x) (1 - 2x) y = [2 + (-x) + (-x)] [1 + (-2x) + (-2.x)] - (2 - 4x - x + 2x2)y = 2 - 4x - 4.x - x + 2x2 + 2x.x - x + 2x.x + 2.(x)2 - 2 + 5x - 2x2.

y = - 5 x + 4x. x + 2.(x)2 Factorando y dividiendo para x : y = x (-5 + 4x + 2 x) x x y lim=

.

- 5 + 4x + 2( 0 )

x 0

dy = - 5 + 4x = 4x - 5 dx 25. y = (Ax + B) (Cx + D) y + y = [A (x + x) + B] [C (x + x) + D]y + y - y=

y - y + y = (Ax + A. x + B ) (Cx + C. x + D ) - (Ax + B) (Cx + D).

[A (x + x) + B] [C (x + x) + D] - (Ax + B) (Cx + D).

y = ACx 2 + ACx.x + ADx + ACx.x + AC(x)2 + AD(x) + BCx + BC.x + BD - ACx 2 - ADx - BCx - BD .

y = 2 ACx. x + AC(x)2 + AD(x) + BC. x Factorando y dividiendo para x. y = x (2Acx + AC. x + AD + BC) x xx

.

y = 2ACx + AC.x + AD + BC = 2Acx + [AC(0)] + AD + BC lim x 0 y = 2ACx + 0 + AD + BC

lim x0

dy = 2ACx + AD + BC dx 26. s = (a + bt)3 s + s = [a + b (t + t)]3 s + s - s = [a + b (t + t)]3 - (a + bt)3 s=

[a + bt + bt]3 - [a3 + 3a2bt + 3a(bt)2 +(bt)3]

s = a3 + (bt) 3 + (bt)3+ 3a2(bt) + 3a2(bt) + 3a(bt) 2+ 3(bt)2(bt) + 3a(bt)2 + 3(bt)(bt)2 + 6a(bt) (bt) - a 3 - 3a2(bt) - 3a(bt) 2 - (bt) 3 s = (bt)3+ 3a2(bt) + 3(bt)2 (bt) + 3a (bt)2 + 3 (bt) (bt)2 + 6a (bt) (bt)

Factorando , dividiendo y simplificando para t . s = t {(bt)2 + 3a2b + 3b3t2 + 3ab2t + 3b3t.t + 6ab2t} t t .t lim t0

s

=

[b (0)2 + 3a2b + 3b3t2 + 3ab2.(0) + 3b3t.(0) + 6ab2t

ds = 0 + 3a2b + 3b3t2 + 0 + 0 + 6ab2t = 3a2b + 3b3t2 + 6ab2t. dt ds = 3a2b + 6ab2t + 3b3t2 = 3b ( a2 + 2abt + b2t2 ) = dt ds = {3b [a + (bt)]2} dt

27.

y=

x . a + bx2 y + y = x + x . 2 a + b (x + x) y + y - y = x + x x . 2 a + b (x + x) a + bx2 y = (x + x) (a + bx2) - x {a + b (x + x)2} [a + b (x + x)2] [a + bx2] y = ax + bx3 + a. x + bx2. x - x{a + b[x2 + 2x.x + (x)2]} [a + b (x + x)2] [a + bx2]y = ax + bx3 + a. x + bx2. x - x{a + bx2 + 2bx.x + b.(x)2} [a + b (x + x)2] [a + bx2]

y = ax + bx3 + a.x + bx2. x - ax - bx3 - 2bx2. x - bx.(x)2. [a + b (x + x)2] [a + bx2]y = a. x - bx2.x - bx.(x)2 . Factorando y dividiendo para x: [a + b (x + x)2] [a + bx2]

lim x0

y = x (a - bx2 - bx.x) . 2 2 x [a + b (x + x) ] [a + bx ] x . y = a - bx2 - bx.x a - bx2 - bx ( 0 ) . = 2 2 2 2 x [a + b (x + x) ] [a + bx ] {a + b [x + ( 0 )] }[a + bx ] y a - bx2 - 0 .

=

x x0

[a + bx2] [a + bx2]

dy = a - bx2 dx [a + bx2]2 28. y = a + bx2 x2 y + y = a + b (x + x)2 (x + x)2 y + y - y = a + b (x + x)2 - [a + bx2] (x + x)2 x2 y = {a + b (x + x)2} (x2) - (x + x)2 (a + bx2) (x + x)2 x2 2 y ={a + b[x +2x.x + (x)2]}(x2) - {x2 + 2x.x + (x)2}(a +bx2) y ={a + bx + 2bx . x + b. (x)2}(x2) - {ax2 + bx4 + 2ax. x (x + x)2 x2 + 2bx3. x + a (x)2 + bx2.(x)2} (x + x)2 x2 2 4 y = ax + bx + 2bx3.x + bx2(x)2 - ax2- bx4-2ax.x (x + x)2 x2 y = 2bx3. x - a(x)2 - bx2.(x)2 (x + x)2 x2 y = - 2ax.x - a(x)2 (x + x)2 x22

(x + x)2 x2

Factorando , dividiendo y simplificando para x :

y = x {-2ax - a (x)} = (x) {-2ax - a (x)} x (x + x)2. x2. (x) (x + x)2. x2. (x) yx lim x0=

{-2ax - a (x)} (x + x)2. x2

=

- 2ax - a ( 0 ) = - 2ax - 0 (x + 0 )2 .x2 x2.x2

dy = - 2ax = - 2a.x dx x 4 x3.x 29. y= x2 . a + bx2

=

- 2a x3

y + y = (x + x)2 . a + b (x + x)2 y + y - y = (x + x)2 x2 . [a + b (x + x)2] (a + bx2)

y = (x + x)2 (a + bx2) - {a + b (x + x)2}( x2) [a + b (x + x)2 ] ( a + bx2 )y = {x2 + 2x. x + (x)2}(a + bx2) - {a + b [x2 + 2x. x + (x)2]}( x2) [a + b (x + x)2 ]( a + bx2 )y = {ax2+bx4+2ax(x)+2bx3(x)+a(x)2+bx2(x)2}-{a+bx2+2bx(x)+b(x)2}(x2) [a + b (x + x)2 ] ( a + bx2)

y = ax2 +bx4+2ax.x+2bx3x +a(x)2 +bx2(x)2- ax2 +bx4 +2bx3.x +bx2(x)2 [a + b (x + x)2 ] ( a + bx2 )

y =

2ax.x + a(x)2 [a + b (x + x)2 ] ( a + bx2 )

.

Factorando , dividiendo y simplificando para x: y = (x) {2ax + a(x)} [a + b (x + x)2]( a + bx2 )(x) y = (x) {2ax + a(x)} [a + b (x + x)2](a + bx2) (x) y x= =

.

.

(2ax + a.x) [a + b (x + x)2 ] ( a + bx2)

=

2ax + a ( 0 ) [a + b (x + 0)2] (a + bx2)

.

lim x0 dy = dx 2ax + 0 (a + bx2) (a + bx2) 2ax (a + bx2)2

=

Problemas - Paginas : 34 y 35Aplicando las Derivadas, hallar la pendiente y la inclinacin de la tangente a cada una de las curvas siguientes en el punto cuya abscisa se indica.

1.

y = x2 - 2 , siendo x = 1. dy = 2x = 2 ( 1 ) = 2 tg = 2 = m. = arc tg 2 = 63o26'5''

2.

y = 2x - 1 x2 , siendo x = 3. 2 dy = 2 - 1 . (2x) = 2 - x dx 2 dy = 2 - x = 2 - (3) = 2 - 3 = - 1 dx m = dy = - 1 dx tg = - 1 =

arc tg - 1 = 135o

3.

y = 4 , siendo x = 2. x-1 dy = (- 4 ) .d(x-1) = (- 4 ) .( 1 ) = (- 4 ). dx (x-1)2 dx (x-1)2 (x-1)2 Sustituyendo x = 2, en y'. dy = (- 4 ) = - 4 = - 4 = - 4 . dx (2-1)2 (1)2 1 m=-4 tg = - 4 = arc tg (- 4) = 104o 2' 10''

4.

y = 3 + 3x - x3 , siendo x = -1 y' = 3 - 3x2 y' = 3 - 3 (-1)2 = 3 - 3 (1) = 3 - 3 = 0 . m = tg 0 = 0 . =

arc tg (0) = 0o

5.

y = x3 - 3x2 , siendo x = 1 y' = 3x2 - 6x. Sustituyendo: x = 1 , en y'. y' = 3 (1)2 - 6(1) = 3 (1) - 6 = 3 - 6 = - 3 . m = tg = - 3 .

= arc tg ( - 3 ) = 108o 26' 5''

6.

Hallar el punto de la curva y=5x - x2 en el que la inclinacin de la tangente es de 45o. y = 5x - x2. Segn dato del problema tg 45 = 1 . y' = 5 - 2x. m = 1 . m = y' = 5 - 2x = 1. solucionando la ecuacin: 5 - 2x = 1 ; 5 - 1 = 2x 2 = x ; x =2 . Sustituyendo x = 2 en la ecuacin original. y = 5x - x2. y = 5 ( 2 ) - ( 2 )2 y = 10 - 4 = 6. y=6. P ( 2 , 6 )

7.

En la curva y = x3 + x hallar los puntos en los que la tangente es paralela a la recta y = 4x. Derivando la "curva" y "la recta": y = x3 + x. y' = 3x2 + 1. m1 = 3x2 + 1. y = 4x. y' = 4 m2 = 4

Cuando 2 rectas son paralelas sus pendientes son iguales. m1 = m2 3x2 + 1 = 4 . Solucionando:

x = 1. En la curva reemplazamos x = 1. y = x3 + x . y1 = (1)3 + (1) y1 = 1 + 1 y1 = 2 P1 (1 , 2) y = x3 + x y2 = (-1)3 + (-1) y2 = -1 -1 = -2 y2 = -2 P2 (-1 , -2)

En cada uno de los siguientes problemas hallar: a) b) 8. Los puntos de intercepcin del par de curvas dado. La pendiente y la inclinacin de la tangente a cada curva, y el ngulo formado por las tangentes en cada punto de intercepcin. y = 1 - x2. y = x2 - 1. Igualamos las 2 curvas. 1 - x2 = x2 - 1 . 1 + 1 = x2 + x2 = 2x2 = 2 x2 = 2/2 ; x2 = 1 ; x = 1 Derivamos cada curva para encontrar sus pendientes: y = 1 - x2. y = x2 - 1. y' = - 2x. y'= 2x Cuando: x = 1 m1 = - 2x m1 = - 2(1) = - 2 m2 = 2x m2 = 2(1) =

m1 = - 2

m2 = 2

tg = m1 - m2 = - 2 - 2 = - 4 1+ m1.m2 1 + (-2) (2) 1 - 4 tg = 4 ; 3 = arc tg 4 = 53 8' 3 Cuando: x = -1 m1 = - 2x = - 2(-1) = 2 m1 = 2 ; tg = m1 - m2 1+ m1.m2=

=

-4 -3

=

4 . 3

m2 = 2x = 2 (-1) = -2 m2 = - 2=

2 - (-2) = 2 + 2 1 + (2) (-2) 1 - 4 = arc tg (- 4/3) = 126 52' 11"

4 = -4 . -3 3

Puntos de intercepcin y = 1 - x2 ; Cuando: x = 1 y = 1 - (1)2 y=1-1 P1 = ( 1, 0 ) y=0 Cuando x = -1 y = 1 - x2 . y = 1 - (-1)2 y=1-1 y=0 P2 = ( -1 , 0 )

9.

y = x2 (1) x - y + 2 = 0 (2) Igualamos las 2 curvas en funcin de ''y'' para encontrar sus intercepciones. y = x2. y = x + 2. (1) (2)

x2 = x + 2 x2 - x - 2 = 0 (x - 2) (x + 1) = 0 x = 2 ; x = -1 Derivamos cada curva para encontrar sus pendientes: y = x2 y=x+2 y' = 2x m1= 2x y' = 1 m2 = 1 Cuando: x = 2 m1 = 2x m1 = 2(2) = 4 m1 = 4 m2 = 1 m2 = 1 m2 = 1 tg = m1 - m2 . 1+ m1.m2=

tg =

4-1 = 3 1 + (4)(1) 1 + 4

3 = 0,6 5

=

arc tg (0,6) = 3057'49"

Cuando: x = - 1 m1 = 2x m1 = 2(-1) = - 2 m1 = - 2 tg = m1 - m2 = - 2 - (1) 1+ m1.m2 1 + (-2) (1) m2 = 1 m2 = 1 m2 = 1 tg =

-2-1 1-2

=

-3 -1

=

3.

= arc tg ( 3) = 71 33' 54" Puntos de intercepcin Cuando x = 2 y = x2 y = (2)2 = 4 . P1 (2 , 4) Cuando x = -1 y = x2 y = (-1)2 = 1 P2 (-1 , 1)

Problemas - Paginas : 44 , 45 y 46Comprobar cada una de las siguientes derivadas. 9. d (3x 4 - 2x2 + 8) = 12x 3 - 4x dx d (3x 4) - d (2x2) + d (8) dx dx dx 3.d (x 4) - 2.d (x2) + 0 dx dx 3 (4x 3) - 2 (2x) = 12x3 - 4x 10. d (4 + 3x - 2x3) = 3 - 6x2. dx d (4) + d (3x) - d (2x3) dx dx dx 0 + 3.d (x) - 2 d (x3) dx dx 3(1) - 2 (3x2) = 3 - 6x2 11. d (at5 - 5bt3) = 5at 4 - 15bt2. dt d (at5) - d (5bt3) = a.d (t5) - 5b.d (t3) dt dt dt dt

a(5t4) - 5b (3t2) = 5at4 - 15bt2 12. d ( z2 - z7) = z - z6. dz 2 7 d (z2) - d ( z7) = 1 d (z2) - 1 d (z7) dz 2 dz 7 2 7 1 (2z) - 1 (7z6) = z - z6 2 7 13. d v = 1 . dv dx 2v dx dv dx 2(v)2-1 14..

.

1 . dv = 1 . dv = 2(v)1 dx 2v dx

d( 2-3)=-2+6. dx x x2 x2 x3 d ( 2 ) - d ( 3 ) = - 2 . dx - ( -3 ) . d ( x2 ) = (-2 ).(1) + 3 (2x) = dx x x2 (x)2 dx ( x2)2 dx x2 x4 = - 2+ 6 x2 x3 d (2t 4/3 - 3t 2/3) = 8 t1/3 - 2t -1/3 dt 3 d (2t4/3) - d (3t2/3) = 2 d (t 4/3) - 3 d (t2/3) dt dt dt dt 2. 4. t 4/3-1 - 3 . 2 . t 2/3-1 = 8 t 1/3 - 2 t -1/3

15.

3 16.

3

.

d (2x 3/4 + 4x -1/4) = 3 x -1/4 - x -5/4 dx 2 d (2x 3/4) + d (4x -1/4) = 2 d (x3/4) + 4 d (x -1/4) dx dx dx dx 2 . 3 . x 3/4-1 + 4 (-1). x -1/4-1 = 3 x -1/4 - x -5/4 4 4 2

17.

d (x2/3 - a2/3) = 2 x -1/3. dx 3 d (x2/3) _ d (a2/3) = 2 x2/3-1 - 0 = 2 x -1/3 dx dx 3 3

18.

d ( a +bx + cx2 ) = c - a . dx x x2 d ( a + bx + c.x.x ) = d ( a ) + d ( b ) + d ( c.x ) dx x x x dx x dx dx (-a). d (x) + 0 + c.d (x) = -a + c (1) = c - a x2 dx dx x2 x2

19.

y = x - 2 2 x

;

dy = 1 + 1 . dx 4x x x

dy = 1 .d (x) - (-2) . d (x) dx 2 dx (x)2 dx dy = 1 .dx/dx + 2 .dx/dx dx 2 2x x 2x 1. 1 + 2. 1 = 1 + 1 2 2x x 2 x 4x xx

=

20.

s = a + bt + ct2 t s = a + bt + ct2 t t t

;

ds = - a + b + 3ct . dt 2t t 2 t 2=

a + b.t2/2 + c.t4/2 = a + b.t1/2 + c.t3/2. t1/2 t1/2 t1/2 t1/2

ds = d ( a ) + d ( b.t1/2) + d ( c.t3/2) dt dt t1/2 dt dt ds = - a . d ( t1/2) + b . d ( t1/2 ) + c. d ( t3/2) dt (t1/2)2 dt dt dt ds = - a . 1 .t1/2-1 + b . 1 . t1/2-1 + c. 3 . t3/2-1 dt t 2 2 2 ds = - a . t -1/2 + b . t -1/2 + 3c. t 1/2 dt 2t 2 2 ds = _ a + b + 3c.t dt 2.t.t 2.t 2 21. y = ax + a . ; ax y = (ax)1/2 + dy = a a . dx 2.ax 2x.ax=

-

a + b + 3c.t1/2 . 2.t.t1/2 2.t1/2 2

a = a1/2.x1/2 + a = a1/2.x1/2 + a2/2 . 1/2 1/2 (ax) a . x1/2 a1/2. x1/2

y = a1/2. x1/2 + a1/2 . x1/2 dy = d (a1/2. x1/2) + d ( a1/2 ) = a1/2 . d (x1/2) + a1/2.d ( x -1/2). dx dx dx x1/2 dx dx

dy = a1/2. 1 . x 1/2-1 + a1/2. - 1 . x -1/2-1 = a1/2. x -1/2 - a1/2. x -3/2 dx 2 2 2 2 dy = a1/2 - a1/2 dx 2x1/2 2x3/2 Multiplicamos y dividimos por a1/2 , a cada sumando : dy = a1/2.a1/2 - a1/2.a1/2 dx 2x1/2.a1/2 2x3/2.a1/2=

a a . 2x1/2.a1/2 2x.x1/2.a1/2

dy = a a a a = . dx 2x.a 2x.x.a 2.ax 2x .ax 22. r = 1 - 2 ; dr = 1 d 1 - 2 .

r = (1 - 2)1/2 dr = d [(1 - 2)1/2] = 1 (1 - 2)1/2-1.d (1 - 2) d d 2 d dr = (1 - 2 ) -1/2.( - 2 ) = - (1 - 2) -1/2 = - 1 = d 2 (1 - 2)1/2 23. f ( t ) = (2 - 3t2)3 ; f '(t) = - 18t(2-3t2)2. 1 . 1 - 2

f '( t ) = 3(2 - 3t2)3-1.d (2-3t2) dt f '( t ) = 3(2 - 3t2)2.(0 - 6t) = 3(2 - 3t2)2(-6t) = -18t (2 - 3t2)2

24.

f (x) = 4 - 9x f '(x) = (4-9x)1/3

;

f '(x) =

-3 . (4 - 9x) 2/3

=

1 (4-9x)1/3-1.d (4-9x) = 1 (4-9x)1/3-1.(0 - 9) = 3 dx 3

=

1 (4-9x)1/3-1 (- 9) = - 3(4-9x) -2/3 = 3 . 3 (4-9x) 2/3 1 a2 - x2 ; dy = x . dx (a2 - x2)3/2

25.

y= y=

2 2 - 1/2 1 . = (a - x ) 2 2 1/2 (a - x ) dy = - 1 (a2 - x2)- 1/2 -1.d (a2 - x2) = - (a2 - x2)- 3/2 .(0 - 2x) . dx 2 dx 2

dy = -1 . (- 2.x) = x . dx 2 (a2 - x2) 3/2 (a2 - x2) 3/2 26. f () = (2 - 5)3/5 ; f '() = 3 . 2/5 (2 - 5)

f '() = 3 (2 - 5)3/5-1 . d (2 - 5) 5 d f '() = 3(2 - 5 )-2/5 (0 - 5) = 3 (- 5 ) = - 3 . 2/5 2/5 5 5 (2 - 5) (2 - 5)

27.

y=a-b x

2

;

dy = 2b a - b dx x2 x

.

dy = 2 ( a- b )2-1.d (a - b ) dx x dx x dy = 2 a - b . d (a) - d ( b ) dx x dx dx x dy = 2 a - b [- (-b.x -1-1)] dx x dy = 2 a - b [b.x-2] = 2 a - b dx x x 28. y= a+b x23 =

2 a-b x

0 - d (b.x -1) dx

b x2

=

2b a - b . x2 x

; y' = - 6b x33-1

a+b 2. x2 .

y'= 3 a + b x2

.d a+b dx x2

y'= 3 a + b 2 . -b . d (x2) x2 (x2)2 dx y'= 3 a + b 2 . -b (2x) = - 6b a + b 2 x2 x4 x3 x2

29.

y = x a + bx y = x (a + bx)1/2

; y' = 2a + 3bx . 2(a + bx)1/2

y'= x.d (a + bx)1/2 + (a + bx)1/2.d (x) dx dx y'= x. 1 .(a + bx)1/2-1.d (a + bx) + (a + bx)1/2(1) 2 dx y'= x(a + bx)-1/2(b) + (a + bx)1/2 = bx + (a + bx)1/2 2 2(a + bx)1/2 y'= bx + 2(a + bx)1/2.(a + bx)1/2 = bx + 2(a + bx) = bx + 2a + 2bx. 2(a + bx)1/2 2(a + bx)1/2 2(a + bx)1/2 y'= 2a + 3bx . 2(a + bx)1/2 ; s'= a2 + 2t2 a2 + t2

30.

s = t a2 + t2 s = t (a2 + t2)1/2

ds = t.d (a2 + t2)1/2 + (a2 + t2)1/2.dt dt dt dt ds = t. 1 .( a2 + t2)1/2-1.d (a2 + t2) + (a2 + t2)1/2.( 1 ) dt 2 dt 2 2 -1/2 ds = t( a + t ) .( 2.t ) + ( a2 + t2)1/2 = t2 + ( a2 + t2)1/2 . dt 2 ( a2 + t2)1/2 ds = t2 + {(a2 + t2)1/2}2=

t2 + a2 + t2 = a2 + 2t2

.

( a2 + t2)1/2 31. y= a-x a+x dy = dx ; y'= -

( a2 + t2)1/2 ( a2 + t2) 2a . (a + x)2

(a+x).d (a-x) - (a-x).d (a+x) dx dx = (a + x) ( -1 ) - ( a - x) ( 1 ) 2 (a + x) (a + x)2=

dy = - a - x - a + x dx (a + x)2 32. y = a2 + x2 a2 - x2 dy = dx ;

_

2a . (a + x)2 4a2x .

y' =

(a2 - x2)2

(a2 - x2).d (a2 + x2) - (a2 + x2).d (a2 - x2) dx dx . 2 2 2 (a - x )

dy = (a2 - x2) (2x) - (a2 + x2) (- 2x) = 2a2x - 2x3 + 2a2x + 2x3 dx (a2 - x2)2 (a2 - x2)2 dy = 4a2x . dx (a2 - x2)2 33. y = a2 + x2 x y = (a2 + x2)1/2 x x.d (a2 + x2)1/2 - (a2 + x2)1/2.d (x) y'= dx dx . ; y' = - a2 . x2 a2 + x2

x2 x. 1 . (a2 + x2)1/2-1.d (a2 + x2) - (a2 + x2)1/2(1) y'= 2 dx . x2 x(a2 + x2)-1/2(2x) - (a2 + x2)1/2 x(a2 + x2)-1/2(2x) - (a2 + x2)1/2 y'= 2 2 . = 2 2 x x x2 - (a2 + x2)1/2 y'= (a2 + x2)1/2 x2 x2 - (a2 + x2)1/2. (a2 + x2)1/2 (a2 + x2)1/2 . x2.

=

x2 - (a2 + x2) x2 - a2 - x2 -a2 y'= (a2 + x2)1/2 = (a2 + x2)1/2 = (a2 + x2)1/2 x2 x2 x2 1 1 1 y'= - a2 x (a + x2)1/22 2 =

=.

- a2 x a2 + x22

.

34.

y= y=

x a - x22

;

y'= a2 . 2 (a - x2)3/2

x (a - x2)1/22

.

(a2 - x2)1/2.d (x) - x.d {(a2 - x2)1/2} y'= dx dx . {(a2 - x2)1/2}2y'= (a2 - x2)1/2(1) - x. 1 .(a2 - x2)1/2-1.d (a2 - x2)} 2 dx (a2 - x2)2/2

.

(a2 - x2)1/2 - x.(a2 - x2)-1/2(- 2x)} (a2 - x2)1/2 +

x2

.

y'=

2 (a2 - x2)2/2

=

(a2 - x2)1/2 . (a2 - x2)

y'=

(a2 - x2)1/2. (a2 - x2)1/2 + x2. 2 2 1/2 2 2 1/2 (a2 - x2)1/2 + x2 . = (a - x ) . (a - x ) 2 2 2 2 2 2 1/2 (a - x ) (a - x )(a - x )=

y'= (a2 - x2) + x2 (a2 - x2)3/2 35. r = 2 3 - 4 r = 2 .(3 - 4)1/2

a2 - x2 + x2 (a2 - x2)3/2

=

a2 . (a2 - x2)3/2

;

r'= 6 - 102 . (3 - 4)1/2

r'= 2.d (3 - 4)1/2 + (3 - 4)1/2.d ( 2) d d r'= 2. 1 .(3 - 4)1/2-1.d (3 - 4) + (3 - 4)1/2(2) 2 dr'= 2(3 - 4 )-1/2(- 4 ) + (3 - 4)1/2(2) = - 2 2 + (2)(3 - 4)1/2 2 (3 - 4)1/2

r'= - 2 2 + (2)(3 - 4 )1/2.(3 - 4 )1/2 = - 22 + (2 )(3 - 4) (3 - 4)1/2 (3 - 4)1/2 r'= - 22 + 6 - 82 (3 - 4)1/2 36. y= 1 - cx 1 + cx=

6 - 10 2 . (3 - 4)1/2 c . (1 + cx ) 1 - c2x2

; y'= -

y = (1 - cx)1/2 . (1 + cx)1/2 (1 + cx)1/2.d [(1 - cx)1/2] - (1 - cx)1/2.d [(1 + cx)1/2] dy = dx dx . dx [(1 + cx)1/2]2dy = dx (1 + cx)1/2. 1 (1 - cx)1/2-1.d (1-cx) - (1 - cx)1/2. 1 .(1 + cx)1/2-1.d (1 + cx) 2 dx 2 dx . (1 + cx)

(1 + cx)1/2(1 - cx) -1/2( -c) - (1 - cx)1/2(1 + cx) -1/2( c) dy = 2 2 dx (1 + cx ) - c (1 + cx )1/2 _ c (1 - cx )1/2 . dy = 2 (1 - cx)1/2 2 (1 + cx )1/2 dx ( 1 + cx ) -c [(1 + cx)1/2]2 - c [(1 - cx)1/2]2 dy = 2 (1 - cx)1/2 (1 + cx )1/2 . (1 + cx ) - c (1 + cx ) - c ( 1 - cx ) - c - c2x - c + c2x 1/2 1/2 dy = 2 (1 - cx) (1 + cx ) = 2 (1 - cx)1/2 (1 + cx )1/2 dx ( 1 + cx ) ( 1 + cx )1 1=.

.

=

.

dy = - 2c 1/2 dx 2 (1 - cx) (1 + cx )1/2(1 + cx ) dy = -c . 1/2 1/2 dx (1 - cx) (1 + cx ) (1 + cx )

dy = -c -c . = dx ( 1 + cx ) .1 - cx . 1 + cx (1 + cx ).(1 - cx )(1 + cx ) dy = _ c .

dx

(1 + cx ) 1 - c2x2

37.

y=

a2 + x2 a2 - x2

; y'=

2a2x . 2 2 4 4 (a - x ) (a - x )

y = (a2 + x2)1/2 (a2 - x2)1/2 dy = dx (a2 - x2)1/2.d (a2 + x2)1/2 _ (a2 + x2)1/2.d (a2 - x2)1/2 dx dx [(a2 - x2)1/2]2 .

(a2 - x2)1/2.1. (a2 + x2)1/2-1.d (a2 + x2) _ (a2 + x2)1/2.1.(a2-x2)1/2-1.d (a2-x2) dy = 2 dx 2 dx dx (a2 - x2)

.

(a2 - x2)1/2.1.(a2 + x2)-1/2(2x) - (a2 + x2)1/2.1.(a2-x2)-1/2(- 2x) dy = 2 2 . 2 2 dx (a - x ) dy = dx x .(a2 - x2)1/2 + x.(a2 + x2)1/2 (a2 + x2)1/2 (a2-x2)1/2 2 2 (a - x )

=

x{(a2 - x2)1/2}2 + x {(a2 + x2)1/2}2 dy = (a2 + x2)1/2 (a2-x2)1/2 . dx (a2 - x2) x(a2 - x2) + x (a2 + x2) dy = (a2 + x2)1/2 (a2-x2)1/2 dx (a2 - x2) a2x - x3 + a2x + x3 . 2 2 1/2 (a2-x2)1/2 = = (a + x ) 2 (a - x2)

2 a 2x . dy = (a + x ) (a2-x2)1/2 = 2 a2x = 2 2 2 2 2 2 1/2 dx (a - x ) (a2-x2)1/2 . (a - x ) (a + x )2 2 1/2

1 dy = 2 a2x 2 2 dx (a - x ).(a2 + x2)(a2-x2)=

2 a2x (a - x )(a2 + x2) (a2-x2)2 2

=

dy = 2 a2x . 2 2 4 4 dx (a - x ).(a - x ) 38. s=3

2 + 3t 2 - 3t

; s'=

4 . (2 + 3t)2/3(2 - 3t)4/3

s = (2 + 3t)1/3 (2 - 3t)1/3

(2 - 3t)1/3.d (2 + 3t)1/3 _ (2 + 3t)1/3.d (2 - 3t)1/3 ds = dt dt . dt [(2 - 3t)1/3]2ds = dt (2 - 3t)1/3.1. (2 + 3t)1/3-1.d (2 + 3t) _ (2 + 3t)1/3.1. (2 - 3t)1/3-1.d (2 - 3t) 3 dt 3 dt . (2 - 3t)2/3

(2 - 3t)1/3.1. (2 + 3t)-2/3( 3 ) - (2 + 3t)1/3.1. (2 - 3t)-2/3 (- 3 ) ds = 3 3 . 2/3 dt (2 - 3t) ds = dt (2 - 3t)1/3 + (2 + 3t)1/3 (2 + 3t)2/3 (2 - 3t)2/3 (2 - 3t)2/3

=

(2 - 3t)1/3(2 - 3t)2/3 + (2 + 3t)1/3(2 + 3t)2/3

ds = dt ds = dt

(2 + 3t)2/3(2 - 3t)2/3 (2 - 3t)2/3 (2 - 3t)1/3+2/3 + (2 + 3t)1/3+2/3 (2 + 3t)2/3(2 - 3t)2/3 (2 - 3t)2/3

. (2 - 3t)3/3 + (2 + 3t)3/3 . (2 + 3t)2/3(2 - 3t)2/3 . = (2 - 3t)2/3.

(2 - 3t) + (2 + 3t) 2 - 3t + 2 + 3t ds = (2 + 3t)2/3(2 - 3t)2/3 = (2 + 3t)2/3(2 - 3t)2/3 = dt (2 - 3t)2/3 (2 - 3t)2/3 4 . 2/3 4/3 ds = (2+ 3t)2/3(2 - 3t)2/3 = 4 = (2 + 3t) (2 - 3t) dt (2 - 3t)2/3 (2 - 3t)2/3 1 39. y=

2px

; y'= p . y p . (2px)1/2

dy = 1 . (2px)1/2-1.d (2px) = 1 . (2px)-1/2( 2p ) = dx 2 dx 2 Sustituyendo: y = 2px dy = p . dx y 40. y = b a2 - x2 a y = b (a2 - x2)1/2 a ; y'= - b2x . a2y=

(2px)1/2 , en la derivada.

dy = b . 1 .(a2 - x2)1/2-1.d (a2 - x2) = b . 1 .(a2 - x2)-1/2 ( - 2x ) dx a 2 dx a 2 dy = - bx . Multiplicamos y dividimos por: "a . b" .

dx a (a2 - x2)1/2 dy = - b.x.a.b. - b2x .a. . = dx a.a.b.(a2 - x2 )1/2 a2.b.(a2 - x2 )1/2 Segn el problema: y = b ( a2 - x2 )1/2 ; 1 = a . 2 2 1/2 a y b (a - x ) dy = - b2x . a dx a2 b (a2 - x2 )1/2 41. y = (a2/3 - x2/3)3/2 ;=

- b2x . 1 a2 y y'= 3

=

_ b2x . a2y

y . x y'= 3 . (a2/3 - x2/3)3/2-1.d (a2/3 - x2/3) = 3 . (a2/3 - x2/3)1/2(0 - 2x2/3-1) 2 dx 2 3 y'= 3 . (a2/3 - x2/3)1/2( - 2x2/3-1) 2 3 y'= - (a2/3 - x2/3)1/2 . Elevando al cubo y sacando raiz cbica x1/3 tanto al nmerador y denominador. y'=3

-(a2/3 - x2/3)1/2 x1/3

3

=

3

-(a2/3 - x2/3)3/2 . x3/3

Pero: y = (a2/3 - x2/3)3/2,sustituimos en y'. y'= 3

y x

.

Hallar la derivada de cada una de las siguientes funciones: 42. f (x) = 2x + 3x f (x) = (2x)1/2 + (3x)1/3 . ; f '(x) = 1 + 1 (2x)1/2 (3x)2/3.

f '(x) = 1 . (2x)1/2-1.d (2x) + 1 . (3x)1/3-1.d (3x) 2 dx 3 dxf '(x) = (2x)-1/2(2) + (3x)-2/3(3) = (2x)-1/2 + (3x)-2/3 = 1 + 1 . 2 3 (2x)1/2 (3x)2/3

43.

y= 2-x 1 + 2x2

;

y'= 2x2 - 8x - 1 . (1 + 2x2)2

(1 + 2x2).d (2 - x) - (2 - x).d (1 + 2x2) y'= dx dx . 2 2 (1 + 2x ) y'= (1 + 2x2)( - 1) - (2 - x)(4x) (1 + 2x2)2 y'= - 1 - 2x2 - 8x + 4x2 (1 + 2x2)2 44. y= y= x a - bx x (a - bx)1/2= = =

- (1 + 2x2) - (2 - x)(4x) (1 + 2x2)2

2x2 - 8x - 1 . (1 + 2x2)2

;

y'=

2a - bx . 2(a - bx)3/2.

x . (a - bx)-1/2

y'= x .d (a - bx)-1/2 + (a - bx)-1/2.d (x) dx dx y'= x. - 1 . (a - bx)-1/2-1.d (a - bx) + (a - bx)-1/2(1) 2 dx y'= - x(a - bx)-2/2(a - bx)-1/2 ( - b) + (a - bx)-1/2 2 y'= bx(a - bx)-2/2(a - bx)-1/2 + 2(a - bx)-1/2

2 1 bx + 2 . y'= (a - bx)-1/2{bx(a - bx)-2/2 + 2} = (a - bx)1/2 (a - bx) . 2 2

1 bx +2(a-bx) bx + 2a - 2bx y'= (a - bx)1/2 (a - bx) = (a - bx)1/2(a - bx)2/2 2 2 1 45. s = a + bt t s = (a + bt)1/2 t ds = dt ds = dt ds = dt ds = dt t.d (a + bt)1/2 - (a + bt)1/2.d (t) dt dt t2 . ; s'= - (2a + bt) . 2t2(a + bt)1/2

. =

2a - bx . 2(a - bx)3/2

t. 1 .( a + bt)1/2-1.d (a + bt) - (a + bt)1/2.(1) 2 dt . t2 t. ( a + bt)-1/2(b) - (a + bt)1/2 2 t2 bt - 2{(a + bt)1/2}2 2(a + bt)1/2 = t2 - 2a - bt . bt - (a + bt)1/2. 1/2 2(a + bt) . t2 bt - 2a - 2bt 2(a + bt)1/2. t2

=

bt - 2(a + bt) 2(a + bt)1/2 t2

=

ds = 2(a + bt)1/2 dt t2

=

- 2a - bt = - (2a + bt) . 2t2(a + bt)1/2 2t2(a + bt)1/2

46.

r= 3 a + b r = (a + b)1/3 dr = d dr = d

; r'=

- (3a + 2b ) . 3 2 (a + b )2/3

.d (a + b)1/3 - (a + b)1/3.d () d d . 2 1/3-1 . 1 . (a + b) .d (a + b) - (a + b)1/3. (1) 3 d . 2

.(a + b )-2/3(b) - (a + b)1/3 b - (a + b)1/3 2/3 dr = 3 . = 3(a + b ) 2 2 d b - 3(a + b)1/3.(a + b)2/3 b - 3(a + b ) b - 3a - b 2/3 2/3 dr = 3(a + b)2/3 = 3(a + b ) = 3(a + b ) d 2 2 2 dr = - 3a - 2b d 3 2 (a + b)2/3 47. y = x2 5 - 2x ;=

- (3a + 2b ) . 3 2 (a + b)2/3 y'= 10x - 5x2 . (5 - 2x)1/2

y = x2(5 - 2x)1/2 dy = x2.d (5 - 2x)1/2 + (5 - 2x)1/2.d (x2) dx dx dx

dy = x2. 1 . (5 - 2x)1/2-1.d (5 - 2x) + (5 - 2x)1/2( 2x ) dx 2 dxdy = x2(5 - 2x)-1/2(- 2 ) + (5 - 2x)1/2( 2x ) dx 2=

- x2 + (5 - 2x)1/2( 2x ) (5 - 2x)1/2

dy = - x2 + (5 - 2x)1/2. (5 - 2x)1/2( 2x ) = - x2 + 2x(5 - 2x) dx (5 - 2x)1/2 (5 - 2x)1/2 dy = - x2 + 10x - 4x2 = 10x - 5x2 . dx (5 - 2x)1/2 (5 - 2x)1/248. y = x. 2 + 3x ; y'= 2(2 x + 1). (2 + 3x)2/3

y = x ( 2 + 3x )1/3 dy = x.d ( 2 + 3x )1/3 + ( 2 + 3x )1/3.d (x) dx dx dx dy = x. 1 . (2 + 3x)1/3-1.d (2 + 3x) + (2 + 3x)1/3(1) dx 3 dxdy = x.( 2 + 3x )-2/3( 3 ) + ( 2 + 3x )1/3 = x + ( 2 + 3x )1/3 2/3 dx 3 ( 2 + 3x )

dy = x + ( 2 + 3x )1/3.( 2 + 3x )2/3 = x + ( 2 + 3x ) = x + 2 + 3x dx (2 + 3x)2/3 (2 + 3x)2/3 (2 + 3x)2/3

.

dy = 4x + 2 = 2 (2 x + 1 ). dx ( 2 + 3x )2/3 ( 2 + 3x )2/3

49.

s=

2t - 1 t21/2

; s'= ;1/2-1

(t3 + 1) . t . ( 2t3 - 1 )1/22

s = 2t - 1 t2

ds = d 2t - 1 1/2 dt dt t2 .d 2t - 1 dt t2=

ds = 1 2t - 1 dt 2 t2

1 2t - 1 2 t2

-1/2

.d [(2t - t -2 )] dt

ds = 1 . ( 2t - t-2 )-1/2.[2 -(-2.t -2-1 )] = . 1 .[2 + 2t -3 ] dt 2 2( 2t - t -2 )1/2 2+ 2. 2t3 + 2 3 ds = (2 + 2t ) = t t3 = dt 2( 2t - t -2 )1/2 2 2t - 1 1/2 2( 2t3 - 1 )1/2 t2 ( t2 )1/2-3

2(t3 + 1) t3 = = 2( 2t3 - 1 )1/2 t

.

ds = 2 . t .(t3 + 1) dt 2 . t 3. ( 2t3 - 1 )1/2

=

(t3 + 1) . t . ( 2t3 - 1 )1/22

50.

y = ( x + 2 )2 x2 + 2 ; y = ( x + 2 )2. ( x2 + 2 )1/2

y'= 3x3 + 6x2 + 8x + 8 . (x2 + 2)1/2

dy = ( x + 2 )2.d ( x2 + 2 )1/2 + ( x2 + 2 )1/2.d ( x + 2 )2 dx dx dx

dy = (x + 2)2. 1 . (x2 + 2)1/2-1.d (x2 + 2) + (x2 + 2)1/2.2(x + 2)2-1.d (x + 2) dx 2 dx dx

dy = (x + 2)2(x2 + 2)-1/2.( 2 x ) + (x2 + 2)1/2.2(x + 2)(1) dx 2 .

dy = x (x + 2)2 + 2(x2+2)1/2(x+2) = dx (x2 + 2)1/2 dy = x(x + 2)2 + 2(x2 + 2)1/2.(x2 + 2)1/2.(x + 2) dx (x2 + 2)1/2 dy = x (x2 + 2x + 4) + 2( x2 + 2)(x + 2) = dx (x2 + 2)1/2 dy = x3 + 2x2 + 4x + 2(x3 + 2x2 + 2x + 4) dx (x2 + 2)1/2 dy = x3 + 2x2 + 4x + 2x3 + 4x2 + 4x + 8 dx (x2 + 2)1/2 51. y=

3x3 + 6x2 + 8x + 8 (x2 + 2)1/2

=

1 + 2x . ; y'= x . 1/2 4/3 1 + 3x (1 + 2x) (1 + 3x)

y = ( 1 + 2x )1/2 ( 1 + 3x )1/3dy = dx ( 1 + 3x )1/3.d ( 1 + 2x )1/2 - ( 1 + 2x )1/2.d ( 1 + 3x )1/3 dx dx . [( 1 + 3x )1/3]2

(1 + 3x)1/3. 1 .(1 + 2x)1/2-1.d(1 + 2x) -(1 + 2x)1/2.1.(1 + 3x)1/3-1.d (1 + 3x) dy = 2 dx 3 dx . dx ( 1 + 3x )2/3

(1 + 3x)1/3(1 + 2x)-1/2( 2 ) _ (1 + 2x)1/2(1 + 3x)-2/3( 3 )

dy = dx

2 ( 1 + 3x )2/3

3

.

(1 + 3x)1/3 _ (1 + 2x)1/2 dy = (1 + 2x)1/2 (1 + 3x)2/3 dx ( 1 + 3x )2/3

=

(1 + 3x)1/3(1 + 3x)2/3 - (1 + 2x)1/2(1 + 2x)1/2 dy = (1 + 2x)1/2 (1 + 3x)2/3 = 2/3 (1 + 3x) (1+3x) - (1+2x) dy = (1 + 2x)1/2(1 + 3x)2/3 dx ( 1 + 3x )2/31/2

=

1 + 3x - 1 - 2x (1 + 2x)1/2(1 + 3x)2/3 ( 1 + 3x )2/3

=

x . 2/3 dy = (1 + 2x) (1 + 3x) = x =. dx ( 1 + 3x )2/3 (1 + 2x)1/2(1 + 3x)2/3(1 + 3x)2/3 dy = x 1/2 dx (1 + 2x) (1 + 3x)4/3

=

En cada uno de los siguientes ejercicios, hallar el valor dado de x. 52. y = ( x2 - x )3 ; x = 3 . dy = 3 ( x2 - x )3-1. d ( x2 - x ) dx dx dy = 3 ( x2 - x )2 (2x - 1 ) ; Sustituyendo x = 3. dx dy = 3 [ (3)2 - 3 ]2 [2(3) - 1 ) ] dx

dy = 3 (9 - 3)2.(6 - 1) dx

=

3(6)2.(5)

=

3.(36)(5) = 540

53.

y

=

x + x

; x = 64.

y = ( x )1/3 + ( x )1/2 dy = 1 . ( x )1/3-1.dx + 1 . ( x )1/2-1.dx dx 3 dx 2 dx dy = ( x )-2/3( 1 ) + ( x )-1/2( 1 ) dx 3 2 Cuando x = 64. dy = 1 + 1 2/3 dx 3(64) 2(64)1/2 dy = 1 + 1 12/3 dx 3(2) 2(2)6/2 dy = 1 + 1 dx 48 16 54.= = =

1 + 1 . 3( x )2/3 2( x )1/2

1 + 1 . 6 2/3 6 1/2 3(2 ) 2(2 ) 1 + 1 3(2)4 2(2)3=

=

1 + 1 . 3(16) 2(8)

1+3 48

=

4 = 1 . 48 12

y = (2x)1/3 + (2x)2/3 ; x = 4 dy = d (2x)1/3 + d (2x)2/3 dx dx dx dy = 1 . (2x)1/3-1.d (2x) + 2 (2x)2/3-1.d (2x) dx 3 dx 3 dx

dy = (2x)-2/3(2) + 2 ( 2x )-1/3(2) = 2 + 4 . 2/3 1/3 dx 3 3 3(2x) 3(2x) dy = 2 + 4 2 + 4 = = dx 3(2.4)2/3 3(2.4)1/3 3(8)2/3 3(8)1/3 dy = 2 + 4 dx 3(23)2/3 3(23)1/3 dy = 2 + 4 dx 12 6 55. y = 9 + 4x 2 y = (9 + 4x2)1/2 ; x = 2 dy = 1 .( 9 + 4x2 )1/2-1.d (9 + 4x2) dx 2 dx dy = (9 + 4x2)-1/2 ( 8 x) = dx 2 dy = 4(2) dx ( 9 + 4.22 )1/2 56. y= y= 1 . 25 - x22 -1/2 1 . = ( 25 - x ) ( 25 - x2 )1/2 = = =

2 + 4 3(22) 3(2)=

=

2 + 8 = 10 12 12 12

5 . 6

4x . Cuando x = 2 . ( 9 + 4x2 )1/2 4(2) 8 8 = = 1/2 1/2 ( 9 + 16 ) ( 25 ) 25=

8. 5

dy = - 1 (25 - x2 )-1/2-1.d (25 - x2 ) = - (25 - x2 )-3/2( - 2x) dx 2 dx 2

=

dy = +x . Cuando x = 3 2 3/2 dx (25 - x )dy = 3 3 3 = 3 = = dx (25 - 32 )3/2 (25 - 9 )3/2 (16)3/2 (24)3/2=

3 212/2

=

3 26

=

3 . 64

57.

y = 16 + 3x x y = (16 + 3x )1/2 x dy = dx dy = dxdy = dx dy = dx

; x=3

x. d (16 + 3x )1/2 - (16 + 3x )1/2. d ( x ) dx dx . x2 x. 1 . (16 + 3x )1/2-1.d (16 + 3x ) - (16 + 3x )1/2 ( 1 ) 2 dx . x2x (16 + 3x)-1/2 ( 3 ) - (16 + 3x )1/2 2 x2 3x - 2(16 + 3x)1/2(16 + 3x)1/2 2(16 + 3x)1/2 x2 3x - (16 + 3x )1/2 1/2 2(16 + 3x) . x2 3x - 32 - 6x 2(16 + 3x)1/2 x2 . 1

=

=

3x - 2 (16 + 3x) 2(16 + 3x)1/2 x2

=

dy = - 32 - 3x . ; Sustituyendo: x = 3 en: dy/dx . 2 dx 2x (16 + 3x)1/2dy = - 32 - 3(3) - 32 - 9 = dx 2 (3)2 [(16 + 3(3)]1/2 2.9[16 + 9]1/2=

- 41 = - 41 18(25)1/2 18(5)

=

- 41 . 90

58.

y = x 8 - x2 ; x = 2 y = x (8 - x2)1/2 dy = x. d (8 - x2)1/2 + (8 - x2)1/2. d (x) dx dx dx dy = x. 1 .(8 - x2 )1/2-1.d (8 - x2) + (8 - x2)1/2(1) dx 2 dx 2 -1/2 dy = x(8 - x ) (-2x) + (8 - x2)1/2 = - 2 x2 + (8 - x2)1/2 2 1/2 dx 2 2 .(8 - x ) dy = - x2 + (8 - x2)1/2(8 - x2)1/2 dx (8 - x2)1/2=

- x2 + (8 - x2) (8 - x2)1/2

=

- x2 + 8 - x2 (8 - x2)1/2

dy = 8 - 2 x2 . Cuando x = 2 dx (8 - x2)1/2 dy = [8 - 2(2)2] dx (8 - 22)1/2 59.=

8 - 2(4) = 8 - 8 (8 - 4)1/2 ( 4 )1/2

=

0 2

=

0.

y = x2 1 + x3 ; x = 2 y = x2 (1 + x3)1/2 dy = x2. d (1 + x3)1/2 + (1 + x3)1/2.d (x2) dx dx dx

dy = x2. 1 . (1 + x3)1/2-1.d (1 + x3) + (1 + x3)1/2 (2x) dx 2 dxdy = x2(1 + x3)-1/2 ( 3x2 ) + (1 + x3)1/2(2x) dx 2=

3 x4 + 2x.(1 + x3)1/2 2(1 + x3)1/2

dy = 3x4 + 2x.2 (1 + x3)1/2 (1 + x3)1/2 dx 2(1 + x3)1/2 dy = 3x4 + 4x + 4x4 = 7 x4 + 4 x . dx 2(1 + x3)1/2 2(1 + x3)1/2 Sustituyendo: x = 4 en y'. dy = 7.(2)4 + 4(2) dx 2(1 + 23)1/2 60.=

=

3x4 + 4x ( 1 + x3 ) 2(1 + x3)1/2

7 ( 16 ) + 8 2( 9 )1/2

=

112 + 8 = 120 2( 3 ) 6

=

20 .

y = (4 - x2)3 ; x = 3 dy = 3(4 - x2)3-1.d (4 - x2) dx dx dy = 3(4 - x2)2(- 2x) dx dy = - 6(3) (4 - 32)2 dx=

- 6x (4 - x2)2 . Sustituyendo: x = 3 en y'.

=

- 18 (4 - 9)2

=

- 18(- 5)2 = -18(25) = - 450

61.

y = x2 + 2 2 - x2 (2 - x2).d (x2 + 2) - (x2 + 2).d (2 - x2) dy = dx dx . 2 2 dx (2 - x )

dy = (2 - x2)( 2x ) - (x2 + 2)( -2x ) = (2 - x2)(2x) + (x2 + 2)(2x) dx (2 - x2)2 (2 - x2)2

dy = 2x [ 2 - x2 + (x2 + 2 )] dx (2 - x2)2 dy = 2x ( 4 ) dx (2 - x2)2 dy = 8( 2 ) dx ( 2 - 22 )2 62. y = 5 - 2x ; 2x + 1 y = (5 - 2x)1/2 2x + 1=

=

2x ( 2 - x 2 + x 2 + 2 ) (2 - x2)2

8x . Sustituyendo: x = 2 en y'. (2 - x2)2 16 = 16 ( 2 - 4 )2 ( - 2 )2 x= 1 . 2=

=

16 4

=

4.

(2x + 1). d (5 - 2x)1/2 - (5 - 2x)1/2.d (2x + 1) dy = dx dx . 2 dx (2x + 1) (2x + 1). 1 .(5 - 2x)1/2-1.d ( 5 - 2x) - (5 - 2x)1/2(2) dy = 2 dx . dx (2x + 1)2(2x + 1)( 5 - 2x)-1/2( - 2 ) - 2 (5 - 2x)1/2 dy = 2 dx (2x + 1)2=

- ( 2x + 1) - 2 (5 - 2x)1/2 (5 - 2x )1/2 . (2x + 1)2

- ( 2x + 1) - 2 (5 - 2x)1/2. (5 - 2x)1/2 - 2x - 1 - 2(5 - 2x) dy = (5 - 2x)1/2 (5 - 2x)1/2 . = 2 2 dx (2x + 1) (2x + 1) . 1 1

-2x - 1 - 10 + 4x dy = (5 - 2x )1/2 = 2x - 11 ; Cuando x = 1 . dx (2x + 1)2 (5 - 2x )1/2 (2x + 1)2 2 12 . 1 . - 11 dy = 2 dx {5 - 2.1}1/2{ 2 . 1 . + 1}2 2 2 1 - 11 - 10 = (5 -1)1/2(1+1)4/2 (4)1/2(2)4/2.

=

=

- 10 (22)1/2(2)2

.

dy = - 10 = - 10 = - 10 dx (22/2)(4) (2)(4) 8 63. y = x (3 + 2x) ; x = 3 y = x (3 + 2x)1/2

=

-5 4

.

dy = x.d (3 + 2x)1/2 + (3 + 2x)1/2.d (x) dx dx dx dy = x . 1 . (3 + 2x)1/2-1.d (3 + 2x) + (3 + 2x)1/2 (1) dx 2 dx dy = x (3 + 2x)-1/2( 2 ) + (3 + 2x)1/2 = x + (3 + 2x)1/2 dx 2 ( 3 + 2x )1/2 1/2 1/2 dy = x + (3 + 2x) .(3 + 2x) = x + (3 + 2x) = x + 3 + 2x . dx ( 3 + 2x )1/2 ( 3 + 2x )1/2 ( 3 + 2x )1/2 dy = 3 + 3x ; Cuando x = 3 1/2 dx ( 3 + 2x )

dy = 3 + 3(3) dx ( 3 + 2.3 )1/2

=

3+9 (3 + 6)1/2

=

12 91/2

=

12 3

=

4.

64.

y=

4x + 1 5x - 1

; x=2

y = (4x + 1)1/2 (5x - 1)1/2 dy = dx (5x - 1)1/2.d (4x + 1)1/2 - (4x + 1)1/2.d (5x - 1)1/2 dx dx . [(5x - 1)1/2]2

(5x -1)1/2. 1 .(4x+1)1/2-1.d (4x +1) - (4x +1)1/2. 1 .(5x -1)1/2-1.d (5x -1) dy = 2 dx 2 dx . dx (5x - 1)

(5x - 1)1/2(4x + 1)-1/2( 4 ) - (4x + 1)1/2.(5x - 1)-1/2( 5 ) dy = 2 2 . dx (5x - 1)4(5x - 1)1/2 - 5(4x + 1)1/2 4(5x - 1)1/2(5x - 1)1/2 - 5.(4x + 1)1/2(4x + 1)1/2 dy = 2(4x +1)1/2 2(5x-1)1/2 = 2(4x +1)1/2(5x-1)1/2 . dx (5x - 1 ) (5x - 1 ) 4(5x - 1) - 5(4x + 1) 20x - 4 - 20x - 5 -9 . dy = 2(4x + 1)1/2(5x - 1)1/2 = 2(4x + 1)1/2 (5x - 1)1/2 = 2(4x + 1)1/2(5x - 1)1/2 dx (5x-1) (5x-1) (5x-1)1/2(5x-1)1/2 . 1

dy = -9 -9 . = dx 2(5x - 1)(4x +1)1/2(5x-1)1/2 2(5x - 1)[(4x +1)(5x-1)]1/2 Cuando x = 2.

dy = -9 dx 2(5.2 - 1) [(4.2 +1)(5.2-1)]1/2 dy = -9 dx 2( 9 ) [81]1/2 65. y= x2 - 5 10 - x2=

=

-9 2(10 - 1) [ ( 9 )( 9 ) ]1/2=

.

-1 -1 = 2.[81]1/2 2(9)

-1. 18

; x=3

y = (x2 - 5)1/2 (10 - x2)1/2 (10 - x2)1/2.d (x2 - 5)1/2 - (x2 - 5)1/2.d (10 - x2)1/2 dy = dx dx . dx [(10 - x2)1/2]2(10 - x2)1/2. 1 .(x2 -5)1/2-1.d (x2 -5) - (x2 -5)1/2.1.(10 - x2)1/2-1.d (10-x2) dy = 2 dx 2 dx . dx (10 - x2)

(10 - x2)1/2(x2 -5)-1/2( 2 x ) - (x2 -5)1/2(10 - x2)-1/2( - 2 x ) dy = 2 2 . dx (10 - x2)dy = dx x(10 -x2)1/2 + x(x2 -5)1/2 x(10 -x2)1/2(10 -x2)1/2 + x(x2-5)1/2(x2-5)1/2 (x2 -5)1/2 (10 - x2)1/2 = (x2 -5)1/2(10 - x2)1/2 . 2 (10 - x ) (10 - x2)

x(10 -x2) + x(x2-5) 10x - x3 + x3 - 5x dy =(x2-5)1/2 (10 -x2)1/2 = (x2-5)1/2 (10 -x2)1/2 dx (10 -x2) (10 -x2)

=

5x . (x2-5)1/2 (10 -x2)1/2 (10 -x2)

dy = 5x . 2 2 1/2 2 1/2 dx (10 -x ) (10 -x ) (x -5) dy = 5(3)=

; Cuando x = 3 15 .

dx (10 -32)(10 -32)1/2(32-5)1/2 dy = 15 2 1/2 dx ( 1 )( 1 ) ( 4 )1/2=

(10 - 9) (10 - 9)1/2(9 - 5)1/2=

15 (1 )( 1 )( 22 )1/2

15 ( 1 )( 2 )

=

15 . 2

Problemas - Pagina 50Hallar dy para cada una de las siguientes funciones : dx 1. y=

u6 ,

u

=

1 + 2x 1 2 x=

dy = 6u5 du

du = 2 . dx

1 . x

Pero : dy = dy . du dx du dx dy = ( 6u5) 1 dx x 2. y = (2u - u2) , dy = 2 _ 2u . du 2 2u dy = 1 _ 2u . du 2u=

sustituyendo : 6u5 . x u x3 - x

=

du = 3x2 - 1 dx

Sustituyendo estos resultados en : dy = dy . du dx du dx

dy dx

=

1 _ 2u ( 3x2 - 1 ). 2u

3.

y= a-u a+u

;

u= b-x. b+x

(a + u).d (a - u) -(a - u).d (a + u) dy = du du (a + u)2 du .

dy = (a + u)( - 1 ) - (a - u)( 1 ) du (a + u)2 dy = - a - u - a + u = - 2a . du (a + u)2 (a + u)2 (b + x).d (b - x) - (b - x).d (b + x) dx dx . 2 (b + x)=

du = dx

du = (b + x)( - 1 ) - (b - x)( 1 ) = - b - x - b + x dx (b + x)2 (b + x)2 Sustituyendo: dy y du en dy du dx dx dy = - 2a . - 2 b = 4ab . dx (a + u)2 (b + x)2 (a + u)2(b + x)2

- 2b . (b + x)2

4.

y = u a2 - u2 y = u.( a2 - u2 )1/2

;

u = 1 - x2

dy = u.d ( a2 - u2 )1/2 + ( a2 - u2 )1/2.d ( u ) du du du dy = u . 1 . ( a2 - u2 )1/2-1.d ( a2 - u2 ) + ( a2 - u2 )1/2( 1 ) du 2 du dy = u ( a2 - u2 )-1/2( - 2 u ) + ( a2 - u2 )1/2 = dx 2 dy = - u2 + ( a2 - u2 )1/2. ( a2 - u2 )1/2 dx ( a2 - u2 )1/2 dy = - u2 + ( a2 - u2 ) = - u2 + a2 - u2 dx ( a2 - u2 )1/2 (a2 - u2)1/2 u = 1 - x2 = ( 1 - x2 )1/2 du = d ( 1 - x2 )1/2 dx dx du dx= =

a2 - 2 u2 . (a2 - u2)1/2

1 . ( 1 - x2 )1/2-1.d ( 1 - x2 ) = ( 1 - x2 )-1/2( - 2x ) = 2 dx 2

du = - x . dx ( 1 - x2 )1/2

Sustituyendo : dy y du en dy du dx dx . dy = ( a2 - 2 u2 ).( - x ) = - x( a2 - 2 u2 )2/2 = 2 2 1/2 2 1/2 dx ( a - u ) ( 1 - x ) ( a2 - u2 )1/2 ( 1 - x2 )1/2 du = x(2u2 - a2) . 2 2 1/2 2 1/2 dx ( a - u ) (1-x ) 15x = 15y + 5y3 + 3y5 15.d ( x ) = 15.dy + 5.3.y3-1.dy + 3.5.y5-1.dy dx dx dx dx 15 ( 1 ) = 15.dy + 15. y2.dy + 15.y4.dy dx dx dx 15 = 15dy + 15y2.dy + 15y4.dy = 15 dx dx dx 15.dy ( 1 + y2 + y4 ) = 15 dx dy = 6. 15 1 . = 2 4 2 4 15 ( 1 + y + y ) ( 1 + y + y ) x = y + y x = ( y )1/2 + ( y )1/3 dx = 1 . y1/2-1.dy + 1 . y1/3-1.dy dx 2 dx 3 dx 1 = y -1/2. dy + y -2/3. dy 2 dx 3 dx 1 = 1 . dy + 1 .dy 2y1/2 dx 3y2/3 dx

5.

1 = dy ( 1 + 1 ) = 1. dx 2y1/2 3y2/3 dy = dx 1 1 + 1 2y1/2 3y2/3=

3y2/3 2y1/2

1 + 2y1/2 3y2/3

=

2y1/2 . 3y2/3 3y2/3 + 2y1/2

=

dy = 2 y1/2 . 3y2/3 = 6y2/3 . dx y 1/2 (3y1/6 + 2) (3y1/6 + 2) 7. y2=

2px.

2y2-1.dy = 2p.dx dx dx 2y.dy = 2p(1) ; dy = 2 p = p . dx dx 2 y y 8. x2 + y2=

r2

2x + 2y2-1.dy = d (r2) ; 2x + 2y.dy = 0 ; dy = - 2 x = - x . dx dx dx dx 2y y 9. b2x2 + a2y2 = a2b2 d (b2x2) + d (a2y2) = d (a2b2) dx dx dx 2b2x + 2a2y2-1.dy = 0 dx 2a2y.dy = - 2b2x dx dy = - 2b2x dx 2a2y=

.

- b2x . a2y

10.

x + y = a. x1/2 + y1/2=

a1/2

;

d (x1/2) + d (y1/2) dx dx

=

d (a1/2) dx

1 . x1/2-1 + 1 .y1/2-1.dy = 0 2 2 dx 1 + 1 . dy 2 x1/2 2 y1/2 dx

; x -1/2 + y -1/2.dy = 0 2 2 dx .

=

0 ; 1 .dy = - 1 2y1/2 dx 2x1/2

dy = dx 11.

-1 . 2x1/2 = - 2 y1/2 1 . 2 x1/2 2 y1/2=

=

_ y1/2 = x1/2

y x

.

x2/3 + y2/3

a2/3 ; 2 x -1/3 + 2 y -1/3. dy = 0 3 3 dx ; 2 . dy 3 y1/3 dx3=

2 .x2/3-1 + 2 .y2/3-1 = d (a2/3) 3 3 dx 2 + 2 . dy = 0 3 x1/3 3 y1/3 dx - 2 . dy = 3 x1/3 = _ y1/3 dx 3 y1/3 x1/3 2 12. x3 - 3axy + y3=

_

2 . 3 x1/3

=

y x

.

.

0

3x2 - 3a[x.dy + y.dx] + 3y2.dy = 3x2 - 3ax.dy - 3ay + 3y2.dy dx dx dx dx dx 3dy ( y2 - ax ) dx=

3ay - 3x2 ; dy = 3 ay - 3 x2 = 3 ( ay - x2 ) = dx 3( y2 - ax ) 3 ( y2 - ax )

dy = ( ay - x2 ) dx ( y2 - ax ) 13. x3 + 3x2y + y3=

c3

3x2 + 3 [ x2.dy + y.d ( x2 ) ] + 3y2. dy = d ( c3 ) dx dx dx dx 3x2 + 3[x2.dy + 2xy] + 3y2. dy = 3x2 + 3x2.dy + 6xy + 3y2.dy = dx dx dx dx 3.dy ( x2 + y2 ) = - 3x2 - 6xy ; dy = - 3x2 - 6xy dx dx 3( x2 + y2 ) dy = - 3 x ( x + 2y ) dx 3 ( x2 + y2 ) 14. x + 2xy + y = a x + 2.x1/2.y1/2 + y = a dx + 2[x1/2.d (y1/2) + y1/2.d (x1/2) ] + dy = d ( a ) dx dx dx dx dx 1 + 2 [x1/2. 1 . y1/2-1. dy + y1/2. 1 . x1/2-1] + dy = 0 2 dx 2 dx 1 + 2[ x1/2.y -1/2. dy + y1/2.x -1/2] + dy = 0 2 dx 2 dx 1 + 2 x1/2. dy + 2 y1/2 + dy 2 y1/2 dx 2 x1/2 dx 1 + x1/2. dy + y1/2 + dy y1/2 dx x1/2 dx= = = .

_ ( x2 + 2xy ) ( x2 + y2 )

=

_ x ( x + 2y ) ( x2 + y2 )

0

0

dy (1 + x1/2 ) = - 1 - y1/2 dx y1/2 x1/2 dy = dx ( - 1 - y1/2) x1/2 ( 1 + x1/2 ) y1/2 - x1/2 - y1/2 . 1/2 x1/2 + y1/2 ) y1/2 = - ( x 1/2 1/2 1/2 y +x x ( y1/2 + x1/2 ) 1/2 y.

=

=

dy = _ y1/2 = - y dx x1/2 x 15. x2 + a

x y + y2

=

b2=

x2 + a . x1/2 . y1/2 + y2

b2

2x + a [x1/2. d (y1/2) + y1/2.d (x1/2)] + 2y.dy = d ( b2 ) dx dx dx dx 2x + a[x1/2. 1 .y1/2-1.dy + y1/2. 1 . x1/2-1] + 2y. dy = 0 2 dx 2 dx 2x + a [ x1/2. y -1/2. dy + y1/2. x -1/2] + 2y. dy = 0 2 dx 2 dx2x + a [ x1/2 . dy + y1/2 ] + 2y.dy = 2x + a x1/2 . dy + a y1/2 + 2y.dy 2 y1/2 dx 2 x1/2 dx 2 y1/2 dx 2 x1/2 dx

2y.dy + a.x1/2 .dy dx 2.y1/2 dx dy ( 2y + a.x1/2 ) dx 2.y1/2

=

- 2x - a y1/2 . 2x1/2 - 2x - a y1/2 2x1/2 - ( 2x + a y1/2 ) 2x1/2 ( 2y + a.x1/2 ) 2y1/2

=

=

.

- ( 2x.2x1/2 + a y1/2 ) dy = 2 x1/2 1/2 dx 2y.2y + a x1/2 2.y1/2

=

- ( 4 x3/2 + a y1/2 ) 2 x1/2 3/2 4 y + a x1/2 2.y1/2

=

_ y1/2 ( 4 x3/2 + a y1/2 ) x1/2 ( 4 y3/2 + a x1/2 )

16.

x4 + 4x3y + y4

=

20

d (x4) + d (4x3y) + d (y4) = d ( 20 ) dx dx dx dx 4x3 + 4 [x3.dy + y.d ( x3 )] + 4y3.dy dx dx dx 4x3 + 4 [x3.dy + 3x2y] + 4y3.dy = 0 dx dx 4x3 + 4x3.dy + 12x2y + 4y3.dy = 0 dx dx 4x3.dy + 4y3.dy = - 12 x2y - 4x3 dx dx 4dy ( x3 + y3 ) = - 4 x2 (3y + x) dx dy = - 4 x2 (3y + x) dx 4 ( x3 + y3) 17.= =

0

_ x2 (x + 3y ) ( x3 + y3 )

ax3 - 3b2xy + cy3 = 1 3ax2 - 3b2 [ x.dy + y.dx] + 3cy2.dy = d (1) dx dx dx3ax2 - 3b2[x.dy + y] + 3cy2.dy = 3ax2 - 3b2x.dy - 3b2y + 3cy2.dy dx dx dx dx 3cy2.dy - 3b2x.dy = 3b2y - 3ax2 = 3.dy ( cy2 - b2x ) = 3 ( b2y - ax2) dx dx dx

3.dy ( cy2 - b2x ) = 3 ( b2y - ax2) dx

dy = 3 ( b2y - ax2 ) = ( b2y - ax2 ) = ax2 - b2y dx 3 ( cy2 - b2x ) ( cy2 - b2x ) b2x - cy2 18. y + x x y=

6.

( y )1/2 + ( x )1/2 = 6 ( x )1/2 ( y )1/2 x1/2.d (y1/2) - y1/2.d (x1/2) + y1/2.d(x1/2) - x1/2.d (y1/2) = 0 [(x1/2)]2 [ (y1/2)]2x1/2. 1 .y1/2-1.dy - y1/2. 1 .x1/2-1 y1/2. 1 .x1/2-1 - x1/2. 1 .y1/2-1.dy 2 dx 2 + 2 2 dx = 0 x y

x1/2.y -1/2.dy - y1/2.x -1/2 2 dx 2 x

y1/2.x -1/2 - x1/2.y -1/2.dy + 2 2 dx = 0 y

x1/2 .dy - y1/2 y1/2 - x1/2 .dy 1/2 1/2 2y dx 2x + 2x1/2 2y1/2 dx = 0 x y x1/2 .dy y1/2 y1/2 x1/2.dy 1/2 1/2 1/2 2y dx - 2x + 2x - 2y1/2 dx = 0 x x y y . 1 1 1 1 x 1/2 . dy - y1/2 + y 1/2 - x1/2 . dy = 0. 1/2 1/2 1/2 2. x .y dx 2.x .x 2.x . y 2.y1/2.y dx 1 . dy - y1/2 + 1 - x1/2. dy = 0 2x1/2.y1/2 dx 2x3/2 2x1/2.y1/2 2y3/2 dx 1 . dy - x1/2 . dy=

y1/2 -

1

.

2x1/2.y1/2 dx

2y3/2 dx

2x3/2

2x1/2.y1/2

dy ( 1 - x1/2 ) = y1/2 1 . 1/2 1/2 3/2 3/2 1/2 1/2 dx 2x .y 2y 2x 2x .y dy = y1/2 1 2x3/2 2x1/2.y1/2 1 - x1/2 2x1/2.y1/2 2y3/2 y1/2.y1/2 - x2/2 y - x . 3/2 1/2 2x3/2.y1/2 . = 2x .y 2/2 1/2 1/2 y - x .x . y - x . 2x1/2.y3/2 2x1/2.y3/2

=

dy = (y - x ) . 2 x1/2.y3/2 = x1/2.y3/2 = x1/2. y1/2 . y dx 2 x3/2.y1/2.( y - x ) x3/2.y1/2 x1/2. x . y1/2 dy = x1/2. y1/2 . y dx x1/2. x . y1/2=

y. x

Hallar la pendiente de cada una de las siguientes curvas en el punto dado.

19.

x2 + xy + 2 y2

=

28 ; ( 2 , 3 )

2x + [ x.dy + y.dx ] + 4y.dy = d (28) dx dx dx dx 2x + x.dy + y(1) + 4y.dy = 0 = 2x + x.dy + y + 4y.dy dx dx dx dx x.dy + 4y.dy = - 2x - y dx dx dy ( x + 4y ) dx dy dx= =

- (2x + y ) ; En el punto ( 2 , 3 ) _ (4+3) ( 2 + 12 ) _ 7 14 - 1. 2

_ ( 2x + y ) ( x + 4y )=

m = dy dx

_ { 2(2) + 3 } { 2 + 4(3) }

=

=

=

m = dy = _ 1 . dx 2 20. x3 - 3xy2 + y3 = 1 ; ( 2 , - 1 ) 3x2 - 3 [ x.d ( y2 ) + y2.d ( x ) ] + 3y2.dy = d ( 1 ) dx dx dx dx 3x2 - 3[2xy.dy + y2 (1)] + 3y2.dy = 3x2 - 6xy.dy - 3y2 +3y2.dy . dx dx dx dx 3y2.dy - 6xy.dy = 3y2 - 3x2 dx dx 3ydy ( y - 2x ) = 3 ( y2 - x2 ) dx dy = 3 ( y2 - x2 ) = ( y2 - x2 ) . En el punto ( 2 , 3 ) dx 3 y ( y - 2x ) y ( y - 2x ) m = dy = [ (-1)2 - (2)2 ] dx (-1)[ -1 - 2 (2)] 21. 2x + 3y= =

[1-4] = -3 (-1)( -1 - 4) (-1)(- 5)

=

- 3 . 5

5 ; (2 , 3)=

( 2x )1/2 + ( 3y )1/2

5=

1 .( 2x )1/2-1.d (2x) + 1 .( 3y )1/2-1.d (3y) 2 dx 2 dx

d (5) dx

(2x)-1/2.(2) + ( 3y ) -1/2.(3).dy = 1 + 3 .dy = 3 . dy = - 1 . 2 2 dx (2x)1/2 2(3y)1/2 dx 2( 3y)1/2 dx (2x)1/2

_ dy = dx

1 . ( 2x )1/2 3 2( 3y)1/2

=

_ 2( 3y)1/2 . En el punto ( 2 , 3) 3 (2x)1/2

m = dy = _ 2 [ 3 (3) ]1/2 dx 3 [ 2 (2) ]1/2 22. x2 - 2xy - y2=

=

_ 2 ( 9 )1/2 3 ( 4 )1/2

=

_ 2(3) 3(2)

=

- 1.

52 ; ( 8 , 2 )=

x2 - 2.x1/2.y1/2 - y2

52

2x.dx - 2 [x1/2.d (y1/2) + y1/2.d (x1/2)] - 2y.dy = d ( 52 ) dx dx dx dx dx 2x(1) - 2 [x1/2. 1 .(y1/2-1).dy + y1/2. 1 . (x1/2-1).dx ] - 2y.dy = 0 2 dx 2 dx dx 2x - 2 [ x1/2.y -1/2.dy + y1/2.x -1/2.( 1 ) ] - 2y.dy = 0 2 dx 2 dx 2x - 2 [ x1/2. dy + y1/2 ] - 2y.dy = 2x - 2.x1/2.dy - 2.y1/2 - 2y.dy 2y1/2 dx 2x1/2 dx 2y1/2 dx 2x1/2 dx 2x - 2.y1/2 2.x1/2=

2.x1/2.dy + 2y.dy = 0 2.y1/2 dx dx

2x - y1/2 = x1/2. dy + 2y.dy = 2x - y1/2 x1/2 y1/2 dx dx x1/2 1/2 Sacando factor comun : dy .( x + 2y ) = 2x - y1/2 dx y1/2 x1/2 2x - y1/2 2x.x1/2 - y1/2 dy = x1/2 = x1/2 = 1/2 1/2 1/2 dx x + 2y x + 2y.y y1/2 y1/2 2x3/2 - y1/2 x1/2 . 1/2 x + 2y3/2 y1/2

dy = y1/2 ( 2x3/2 - y1/2 ) . En el punto ( 8 , 2 ) dx x1/2(x1/2 + 2y3/2 ) m = dy = (2)1/2 [ 2(8)3/2 - (2)1/2 )=

(2)1/2 [ 2 (23)3/2 - 21/2 ] .

dx (8)1/2 [(8)1/2 + 2(2)3/2 ) m = dy = ( 2)1/2 [ 2 (23)3/2 - 21/2 ] dx 2( 2)1/2 [(23)1/2 + 2(23/2)]=

2(2)1/2 [ (23)1/2 + 2(23/2) ] [2.29/2 - 21/2] 2(23/2 + 22/2.23/2)=

m = 22/2. 29/2 - 21/2 = ( 211/2 - 21/2 ) = [ 21/2( 210/2 - 1)] = 2[23/2(1 + 22/2)] 2[23/2(1 + 2)] 22/2[23/2(3)] m =dy = 21/2( 25 - 1 ) dx 21/2. 21/2. 23/2. 3 23. x3 - axy + 3ay2= =

32 - 1 24/2. 3

=

31 4.3

=

31 . 12

3a3 ; ( a , a )=

3x2.dx - a [ x.dy + y.dx ] + 3a.2y.dy dx dx dx dx

d ( 3a3) dx

3x2 ( 1 ) - a [ x.dy + y ] + 6ay.dy = 3x2 - ax.dy - ay + 6ay.dy = 0 dx dx dx dx 6ay.dy - ax.dy dx dx=

ay - 3x2

a.dy {6y - x} = ay - 3x2 dx dy = a.y - 3x2 . En el punto (a , a) dx a ( 6y - x ) m = dy = a(a) - 3(a)2 dx a(6.a - a) 24.=

a2 - 3a2 a( 5a )

=

- 2a2 5a2

=

-2a2 = _ 2. 5 a2 5

x2 - xxy - 2y2 = 6 ; ( 4 , 1 ) x2 - x.x1/2.y1/2 - 2y2 x2 - x3/2.y1/2 - 2y2= =

6

6

2x.dx - [ x3/2. 1 .y1/2-1. dy + y1/2. 3 . x3/2-1 ] - 4y. dy = d ( 6 ) dx 2 dx 2 dx dx 2x (1) - [ x3/2.y -1/2. dy + y1/2. 3 . x1/2 ] - 4y. dy 2 dx 2 dx=

0

2x - [x3/2 . dy + 3.x1/2.y1/2] - 4y.dy = 2x - x3/2 .dy - 3 x1/2.y1/2 - 4y.dy 2y1/2 dx 2 dx 2y1/2 dx 2 dx

x3/2 . dy + 4y . dy 2y1/2 dx dx dy ( x3/2 + 4y) dx 2y1/2=

=

2x - 3 x1/2.y1/2 2

2x - 3 x1/2.y1/2 2 4x - 3 x1/2.y1/2 1/2 1/2 1/2 2 = (4x - 3 x .y ).y 3/2 1/2 3/2 3/2 x + 8y.y ( x + 8.y ) 2 y1/2

2x - 3 x1/2.y1/2 dy = 2 3/2 dx ( x + 4y ) 2y1/2 En el punto (4,1)

=

dy = {4(4) - 3 (4)1/2.(1)1/2 }.11/2 dx {(4)3/2 + 8(1)3/2}

=

{16 - (3)(2)(1)} (1) {(22)3/2 + (8)(1)}=

=

{16 - 6} (1) . {(26/2) + 8}

dy = ( 10 ) ( 1 ) dx { 23 + 8 } 25.

=

10 8+8

=

10 16

5. 8

Demostrar que las parabolas y2 = 2px + p2 y y2 = p2 - 2px se cortan en ngulo recto. y2 = 2px + p2 1) y2 = p2 - 2px 2) 2px + p2 = p2 - 2px 2px + 2px = p2 - p2 = 0 4px = 0 Sustituyendo x = 0 en 1) y2 = p y = p

x

=

0

P(0,p); P(0,-p) Derivando (2) y2 = p2 - 2px 2y .dy = 0 - 2p.dx dx dx 2y .dy = 0 - 2p.dx dx dy = _ 2p . dx 2y= =

Derivando ( 1) y2 = 2px + p2 2y.dy = 2p.dx + d (p2) dx dx dx 2y.dy = 2p( 1 ) + 0 dx 2y .dy = 2p dx m = dy = 2p dx 2y p. y p . p

- 2p dx

m = dy = _ p . dx y Pero : y = p m = dy = - p dx y m3 = -p p m4 = - p -p=

Pero : y = p m = dy = p dx y m1 = p = + 1 . p m2 = p = - 1. -p=

=

-p . p

-1 . +1.

=

Las 2 parbolas son perpendiculares , osea que se cortan en ngulo recto , porque el producto de sus pendientes es igual a - 1 .y

y2 = p2 - 2px+p

o

x

y2 = 2px + p2

-p

26.

Demostrar que las circunferencias x2 + y2 - 12x - 6y + 25 = 0 y x2 + y2 + 2x + y = 10 , son tangentes en el punto ( 2 , 1 ).Derivando : x2 + y2 - 12x - 6y + 25 = 0 2x + 2y.dy -12.dx - 6.dy +d (25) = 0 dx dx dx dx Derivando : x2 + y2 + 2x + y = 10 2x + 2y.dy + 2.dx + dy = d (10) dx dx dx dx

2x + 2y.dy -12(1) - 6.dy + 0 = 0 dx dx 2x + 2y.dy -12 - 6.dy = 0 dx dx 2y.dy - 6.dy = 12 - 2x dx dx 2 dy ( y - 3 ) = 2 ( 6 - x ) dx dy = 2 ( 6 - x ) = ( 6 - x ) . dx 2 ( y - 3 ) (y-3) En el punto ( 2 , 1 )dy = ( 6 - x ) dx ( y - 3 )=

2x + 2y.dy + 2(1) + dy = 0 dx dx 2x + 2y.dy + 2 + dy = 0 dx dx 2y.dy + dy = - 2x - 2 dx dx dy = ( 2y + 1 ) = - 2 ( x + 1 ) dx dy = - 2 ( x + 1 ) dx ( 2y + 1 ) En el punto ( 2 , 1 )dy = - 2 (2 + 1) = -2 ( 3 ) = - 6 =-2 dx [2(1) + 1] ( 2 + 1 ) 3

(6-2) = 4 . (1-3) -2

m1 = dy = - 2 dx

m2 = dy = - 2 dx

Si sus pendientes son iguales estas curvas son tangentes.

y

(2,1)

x

27.

Bajo que ngulo corta la recta y = 2x a la curva x2 - xy + 2y2 = 28 .y x2 - xy + 2y2 = 28 (2, 4) y = 2x

x (-2,- 4)

x2 - xy + 2y2 = 28 y = 2x Sustituyendo el valor de y = 2x en x2 - x(2x) + 2(2x)2 = 28 x2 - 2x2 + 2(4x2) = 28 x2 - 2x2 + 8x2 = 28 Sustituyendo el valor de x en y = 2x

7x2 = 28 x2 = 28 = 4 7

y = 2(2) = 4 y = 2(-2) = - 4

P1 (2,4) Puntos de intercepcin x= 2 P2 (-2,-4) Derivando cada curva para encontrar sus pendientes: x2 - xy + 2y2 = 28 2x - {x.dy + y.dx } + 4y.dy = d (28) dx dx dx dx 2x - x.dy - y(1) + 4y.dy = 0 dx dx 2x - x.dy - y + 4y.dy = 0 dx dx 4y.dy - x.dy = y - 2x dx dx dy {4y - x} dx dy = y - 2x . dx 4y - x Sustituyendo el punto P (-2,- 4) . dy = - 4 -2(-2) = - 4 + 4 dx 4(- 4) - (-2) - 16 + 2 m1 = dy = 0 dx= =

y = 2x

y - 2x

0 = 0. - 14

tg = m2 - m1 = 2-0 1 + m1.m2 1 + (0)(2) tg = 2 = arc tg(2) . = 630 26' 6'' .

=

2 = 2 1+0

Problemas Adicionales1. El vertice de la parbola y2 = 2px es el centro de una elipse. El foco de la parbola es un extremo de uno de los ejes principales de la elipse, y la parbola y la elipse se cortan en ngulo recto. Hallar la ecuacin de la elipse .y2 = 4px 4x + 2y2 2 =

p

2

0 a

F(p/2,0)

Si y2 = 4px es el doble de : y2 = 2px El lado recto de es 4p El lado recto de sera la mitad 2p Si el lado recto es 2p, por grfico obtenemos que F (p/2,0) El semi eje principal de la elipse es : a = p/2 El centro de la elipse es el origen (0,0) (x-h)2 + (y-k)2 a2 b2=

1

(x-0)2 + (y-0)2 = 1 ; x2 + y2 = 1 a2 b2 a2 b2 Derivando para obtener la pendiente. x2 + y2 = 1 a2 b2 a-2.d (x2) + b-2.d (y2) = d (1) dx dx dx a-2(2x) + b-2.2y.dy = 0 dx

2y.b-2.dy = - 2x.a-2 dx dy = - 2x.a-2 = - b2x . dx 2y.b-2 a2y m1 = - b2x . a2y

Derivando y2 = 2px para obtener m2. 2y.dy = 2p.d (x) dx dx dy = 2p dx 2y=

2y.dy = 2p dx m2 = p . y

p y

Para que la parbola y la elipse se corten en ngulo recto, el producto de sus pendientes tiene que ser igual a -1. ( m1 ) ( m2 ) = - 1 . - b2x a2y p y=

-1. ; Pero: a = p . 2

b2 = -a2.y2 = a2.y2 . -p.x p.x 2 2 a = p . 4

Sustituyendo en el valor de a2 . b2 = p2.y2 2 4 . = p.y p.x 4x

Sustituyendo en que es la ecuacin de la elipse los valores de a2 y b2.

x2 + y2 a2 b2

=

1

4x2 + 4x p2 p

=

1 1 p2

x2 y2 1 + 1 = 1 p2 p.y2 4 4x 4x2 + 4x.y2 p2 p.y2=

4x2 + 4px p2 4x2 + 4px 1

=

=

Pero : 2y2 = 4px , sustituyendo este valor en : 4x2 + 4px 4x2 + 2y2 = p2 (Ecuacin de la elipse ).

=

p2

2. Se traza un circulo de centro (2a,0) con un radio tal que el circulo corta en ngulo recto a la elipse b2x2 + a2y2 = a2b2.Hallar el radio.

Tomamos primero a la elipse y encontramos su pendiente. Derivando : b2x2 + a2y2 = a2b2. 2b2x + 2a2y.dy = d (a2b2) dx dx 2b2x + 2a2y.dy = 0 dx dy = - 2b2x dx 2a2y m1 = - b2x . a2yLuego se toma a la ecuacin del circulo y se obtiene su pendiente,=

- b2x . a2y

cuyo centro es (2a,0).

( x - 2a )2 + ( y - 0 )2 = r2. ( x - 2a)2 + y2 = r2 Derivando: 2 (x-2a) + 2y.dy = d (r2) dx 2 (x-2a) + 2y.dy = 0 dx m2 = dy = - 2(x-2a) = - ( x - 2a ) 2y y Como el circulo corta en ngulo recto a la elipse, tomamos sus pendientes. m1 . m2 = - 1 -b2x a2y -(x - 2a) y=

-1

b2x ( x-2a ) = - 1 a2.y2 2 2 b x - 2ab2x = - a2y2 b2x2 + a2y2 = 2ab2x Tomamos la ecuacin de la elipse: b2 + a2 = a2b2 igualamos y

b2x2 + a2y2 = a2b2 b2x2 + a2y2 = 2ab2x 2ab2x = a2b2 x = a2b2 2ab2=

a 2

.

Como en la ecuacin de la elipse hay 2 incognitas "x" y "y", sustituimos el valor de x = a y encontramos el valor de y. 2 b2x2 + a2y2 = a2b2 b2 a 22

+ a2y2 = a2b2 a2b2

a2b2 + a2y2 4

=

a2y2 = a2b2 - a2b2 4 a2y2 = 3a2b2 4 2 2 2 y = 3a b = 3b2 4a2 4Como nos piden hallar el radio del circulo, sustitituimos: x = a/2 y y2 = 3b2/4 en la ecuacin del circulo de centro ( 2a , 0 ).

( x-2a)2 + y2 = r2

r2 = 9a2 + 3b2 4

a -2a 2 - 3a 22

2

+ 3b2 = r2 4 r2

r=

9a2 + 3b2 4

=

1 9a2 + 3b2 2

+ 3b2 4

=

r = 9a2 + 3b2 2

.

3. Se une un punto cualquiera "p" de una elipse con los focos. Demostrar que estas rectas forman con la normal a la curva en "p" ngulos agudos iguales. Suponiendo la ecuacin de la elipse: b2x2 + a2y2 = a2b2. y Encontrando su pendiente, derivando: P(x,y) 2b2x + 2a2y.dy = 0 dx F'(-c,o) F(c,o) dy = - 2b2x = - b2x . 2 2 dx 2a y ayAhora la pendiente de la Normal sera:

dx-1= -1 m - b2x a2y=

a 2y b2x

=

Normal.

Segn el grfico: Pendiente FP = y - 0 x-c=

y ; Pendiente F'P = y - o = y . x-c x - (-c) x + c

Aplicando la frmula de un ngulo formado por 2 rectas:

tg = m2 - m1 . 1 + m1.m2 Primero para el ngulo . tg = y - a2y x-c b2x 1 + a2y y ..

b2x x-c b2xy - (a2y)(x-c) 2 2 2 2 2 2 tg = (x-c)b2x) = b xy - a xy + a cy = xy (b -a ) + a cy . b2x(x-c) + a2y2 b2x2 - b2xc + a2y2 b2x2 - b2xc + a2y2 (b2x)(x-c) Pero de la ecuacin de la elipse : b2x2 + a2y2 = a2b2 , despejamos a2y2 a2y2 = a2b2 - b2x2 Y segn la relacin de la elipse: a2 = b2 + c2 a2 - b2 = c2 Sustituyendo estos valores y en tg tg = xy (b2 - a2) + a2cy b2x2 - b2xc + a2y2 tg ==

- xy (a2 -b2) + a2cy . b x - b2xc + (a2b2 - b2x2)2 2 =

- xy ( c2) + a2cy b2x2 - b2xc + a2b2 - b2x2

- c2xy + a2cy = cy ( a2 - cx ). a2b2 - b2xc b2 (a2 - cx )

tg = cy . b2 Luego calculando el ngulo : a2y - y . a2y ( x+c)-b2xy tg = b2x x + c = (b2x )(x + c) = a2xy + a2cy - b2xy . 1+ y a2y (x+c)(b2x)+a2y2 b2x2 + b2cx + a2y2 x+c b2x (x+c)(b2x) tg = xy ( a2 - b2 ) + a2cy

b2x2 + b2cx + a2y2 Sustituyendo estos valores y en tg tg = xy (c2) + a2cy b x + b2cx + a2b2 - b2x22 2 =

c2xy + a2cy b2cx + a2b2

=

cy (cx+a2) . b2(cx+a2)

tg = cy . b2 Como : tg = tg = cy . sus ngulos son iguales, = b24. Demostrar que la recta Bx + Ay = AB es tangente a la elipse b2x2 + a2y2 = a2b2.nicamente si se verifica que: B2a2 + A2b2 = A2B2 Para demostrar que la recta es tangente a la elipse, sus pendientes tienen que ser iguales.

Derivamos para clcular la pendiente de Bx + Ay = AB d (Bx) + d (Ay) = d (AB) dx dx dx B.dx + A.dy = B + A.dy = 0 dx dx dx dy = _ B . dx A Pendiente de la elipse: b2x2 + a2y2 = a2b2 d (b2x2) + d (a2y2) = d (a2b2) dx dx dx 2b2x + 2a2y.dy = 0 dx

dy = _ 2b2x dx 2a2y

=

_ b2x . a2y

Igualando ambas pendientes: _ B A b2x a2y= =

_ b2x . a2y

B . A

x = a2By . Sustituyendo en Ab2 b2x2 + a2y2 = a2b2 b2 a2By 2 + a2y2 = a2b2 Ab2 a4b2B2y2 + a2.A2.b4.y2 = a2.b2.A2.b4 a2.b2.y2(a2.B2 + A2.b2) = a2.b2.A2.b4 y2 y==

a2.b2.A2.b4 . 2 2 2 2 a .b (a .B + A .b )2 2

A2.b4 . 2 2 2 2 (a .B + A .b ) 2 2 2 y= A2.b4 A.b2 = = A.b = A.b = b (a2.B2 + A2.b2) (a2.B2 + A2.b2) A2.B2 A.B BComo esta en funcin de "y",entonces reemplazamos el valor de y en "x". x = a2By . Ab2

a2.B.A.b2 .

x = a2B2 + A2b2 A.b2 1

=

a2.B.A.b2 . 2 2 2 2 2 A.b a B + A b

Pero: a2B2 + A2b2 = A2B2, reemplazando en "x". x = a2.B = a2.B = a2 . A2B2 A.B A x = a2 . A Sustituyendo ahora el valor de "x"e "y"en Bx + Ay = AB B a2 + A b2 A B a2B + Ab2 A B= =

AB

AB

a2.B2 + A2.b2 = AB AB a2B2 + A2b2=

A2B2

B2a2 + A2b2 = A2B2 . { L.q.q.d. (Lo que se queria demostrar)}.

Problemas. Pagina 56Hallar las ecuaciones de la tangente y de la normal a las curvas siguientes en el punto dado. 2. y = x3 - 3x ; (2,2)

dy = 3x2 - 3 . Sustituyendo: P(2,2) en la derivada o pendiente. dx m = dy = 3 (2)2 - 3 = 12 - 3 = 9 dx Ecuacin de la Tangente: y - y1 = m (x - x1) y - 2 = 9 (x - 2) y - 2 = 9x - 18 0 = 9x - y - 18 + 2 = 0 9x - y - 16 = 0 Ecuacin de la Normal: y - y1 = _ 1 (x - x1) m1 y - 2 = - 1 (x - 2) 9 9y - 18 = - x + 2 3. x + 9y - 20 = 0 y = 2x + 1 ; (2,5) 3-x (3-x).d (2x+1) - (2x+1).d (3-x) dy = dx dx . dx (3-x)2

dy = (3-x)(2) - (2x+1)(-1) dx (3-x)2 dy = 6 - 2x + 2x + 1 dx (3-x)2 m = dy = 7 . Pero: P (2,5) dx (3-x)2 m= 7 (3 - 2)2=

7 = 7 (1)2 1

=

7.

Ecuacin de la tangente: y - y1 = m (x - x1) y - 5 = 7 (x - 2) y - 5 = 7x - 14 = 0 7x - y = - 9 = 0 Ecuacin de la Normal: y - y1 = - 1 (x - x1) m1 y - 5 = - 1 (x - 2) 7 7y - 35 = - x + 2 x + 7y - 37 = 0 4. 2x2 - xy + y2 = 16 ; (3,2)

Derivando para encontrar la pendiente: 4x - {x.dy + y.dx } + 2y.dy = d (16) dx dx dx dx 4x - x.dy - y ( 1 ) + 2y.dy = 0 dx dx 4x - x.dy - y + 2y.dy = 0 dx dx dy (2y - x ) = y - 4x dx dy = y - 4x . Pero : P (3,2) dx 2y - x m = dy = 2 - 4(3) dx 2(2) - 3=

2 - 12 4-3

=

- 10 1

=

- 10

Ecuacin de la Tangente: y - y1 = m(x - x1) . Sustituyendo: m = - 10 y P(3,2). y - 2 = - 10 (x - 3) y - 2 = - 10x + 30 10x + y - 32 = 0 Ecuacin de la Normal: y - y1 = - 1 (x - x1) m1 y - 2 = - 1 (x - 3 )

-10 -10(y - 2) = - (x - 3) ; -10y + 20 = - x + 3 x - 10y + 17 = 0 5. y2 + 2y - 4x + 4 = 0 2y.dy + 2.dy - 4.dx + d (4) = 0 dx dx dx dx 2y.dy + 2.dy - 4(1) + 0 = 0 dx dx 2y.dy + 2.dy - 4 = 0 dx dx 2.dy ( y + 1) = 4 dx m = dy = 4 2 . Pero: P(1,-2) = dx 2(y + 1) (y + 1) m= 2 = 2 (-2) + 1 -1=

- 2.

Ecuacin de la Tangente: y - y1 = m (x - x1) y - (-2) = - 2 (x - 1) y + 2 = - 2x + 2 2x + y + 2 - 2 = 0 2x + y = 0

Ecuacin de la Normal: y - y1 = - 1 (x - x1) y - (-2) = - 1 (x - 1) -2 y + 2 = 1 (x - 1) 2 2y + 4 = x - 1 = 2y + 4 . x - 2y - 5 = 06. Obtener las ecuaciones de la tangente y de la normal en ( x1 , y1) a la elipse b2x2 + a2y2 = a2b2.

Derivando la curva: b2x2 + a2y2 = a2b2 2b2x + 2a2y.dy = d (a2b2) dx dx 2b2x + 2a2y.dy = 0 dx 2a2y.dy = - 2b2x dx dy = - 2b2x dx 2a2y=

- b2x . a2y

m = - b2x .Pero : m1 = - b2x1 . a2y a2y1 Ecuacin de la Tangente: y - y1 = m(x - x1) en el punto P (x1,y1) y - y1 = - b2x1 (x - x1)

a2y1 a2y1 (y - y1) = - b2x1 (x - x1) a2y1 y - a2y1.y1 = - b2x1.x + b2x1.x1 a2y1 y - a2y12 = - b2x1.x + b2x12 b2x1.x + a2y1.y = a2y12 + b2x12 Pero: b2x2 + a2y2 = a2b2 b2x12 + a2y12 = a2b2 b2x1x + a2y1y = a2b2 Ecuacin de la Normal:y - y1 = - 1 (x - x1) m1 y - y1 = - 1 ( x - x1) -b2x1 a2y1 y - y1 = a2y1 (x - x1) b2x1 b2x1 (y - y1) = a2y1 (x - x1) ; b2x1.y - b2x1.y1 = a2y1.x - a2y1.x1 a2y1.x1 - b2x1.y1 = a2y1.x - b2x1.y x1.y1 = (a2 - b2) = a2.y1.x - b2.x1.y . Ordenando:

a2.y1.x - b2.x1.y = x1.y1 = (a2 - b2)7. Hallar las ecuaciones de la tangente y la Normal, y las longitudes de la sub-

tangente y la sub-normal, en el punto(x1,y1) de la circunferencia x2 + y2 = r2.

Primeramente derivando la curva: x2 + y2 = r2. 2x + 2y.dy = 0 dy = - 2x = - x .Ahora la pendiente en P (x1,y1) dx 2y y m = - x1 . y1 Ecuacin de la Tangente: y - y1 = m (x - x1) y - y1 = - x1 (x - x1) y1 y1 (y - y1) = - x1 (x - x1) y1.y - y1.y1 = - x1.x + x1.x1 y1.y - y12 = - x1.x + x12 x1x + y1y = y12 + x12 Pero: x2 + y2 = r2 x12 + y12 = r2 Como: y12 + x12 = x12 + y12 = r2 x1x + y1y = r2 Ecuacin de la Normal: y - y1 = - 1 (x - x1) m1 y - y1 = - 1 (x - x1)

-x1 y1 y - y1 = y1 (x - x1) x1 x1 (y - y1) = y1 (x - x1) x1.y - x1.y1 = y1.x - y1.x1 y1x - x1y = x1y1 - y1x1 . y1x - x1y = x1y1 - x1y1 y1x - x1y = 0 o tambien : x1y - y1x = 0 Longitud de la sub-tangente:y1 . m1 y1 = - y12 -x1 x1 y1

ordenando:

Longitud de la sub-normal: m1.y1- x1 y1 = - x1. y1 8. Demostrar que la sub-tangente de la parbola y2 = 2px es bisecada por el vrtice, y que la sub-normal es constante e igual a p. Derivando para obtener la pendiente en P(x1,y1)

y2 = 2px

2y.dy = 2p.dx dx dx dy = 2p(1) dx 2y=

p . m = p . y1 y1

Ecuacin de la tangente: y - y1 = m(x - x1) y - y1 = p (x - x1) y1 y.y1 - y1.y1 = p.x - p.x1 y.y1 - y12 = p.x - p.x1 Pero, la ecuacin de la parbola: y2 = 2px y12= 2px1 y.y1 - 2px1 = p.x - p.x1 y.y1 - 2px1 + px1 - p.x = 0 y.y1 - p.x1 - p.x = 0 (ecuacin de la tangente)Luego encontrando la intercepcin de la tangente con el eje "x". y = 0 x = - x1 las coordenadas de T (- x1, 0) Por grfico observamos que las coordenadas de M (x1,0) demostraremos que TO = OM.

TO = {0 -(-x1) + (0 - 0)2} = ( x1 )2 = x1 = TO OM = {(x1 - 0)2 + (0 - 0)2 = (x1)2 = x1 TO=

OM.

Ahora demostraremos que "P" es igual a la sub-normal. Segn grfico: MN = sub-normal. MN = y1.m1 .sabiendo que m = p , m1 = p . y1 y1 MN = y1.m1 = y1.p = p y1Obtener las ecuaciones de la Tangente y la Normal, y las longitudes de la sub-tangente y la sub-normal de cada una de las siguientes curvas en los puntos indicados.

9.

ay = x2 ; (a,a) y = x2 a=

1 .x2 a

dy = 1 (2x) en el punto (a,a) dx a dy = 2 a = 2 . dx a.

Ecuacin de la Tangente: y - y1 = m (x - x1) y - a = 2 (x - a)y - a = 2x - 2a 2x - y - 2a + a = 0 2x - y - a = 0

Ecuacin de la Normal:y - y1 = - 1 (x - x1) y - a = - 1 (x - a) 2 2(y - a) = - (x - a) 2y - 2a = - x + a x + 2y -2a - a = 0 = x + 2y - 3a.

Longitud de la sub-tangente: y1 = a m1 2.

Longitud de la Sub-normal:y1.m1 (a)(2) = 2a.

10.

x2 - 4y2 = 9 ; (5,2)Derivando para obtener la pendiente en P (5,2) 2x - 8y.dy = 0 dx 2x = dy 8y dx=

x = 5 = 5. 4y 4(2) 8

m= 5 . 8 Ecuacin de la Tangente:

y - y1 = m(x - x1) y - 2 = 5 (x - 5) 8 8(y - 2) = 5(x - 5) 8y - 16 = 5x - 25 = 5x - 8y - 25 + 16 = 5x - 8y - 9 = 0 Ecuacin de la Normal:y - y1 = - 1 (x - x1) m1 y - 2 = - 1 (x - 5) 5 . 8 y - 2 = - 8 (x - 5) 5 5(y - 2) = - 8 (x - 5) 5y - 10 = - 8x + 40 8x + 5y - 10 - 40 = 8x + 5y - 50 = 0

Longitud de la sub-tangente: y1 = 2 m1 Longitud de la Sub-normal: m1.y1 = 5 (2) = 10 = 5 . 8 8 4 2 2 9x + 4y = 72 ; (2,3). Derivando para obtener la pendiente en P (2,3) 18x + 8y.dy = d (72) dx dx

11.

18x + 8y.dy = 0 dx 9 m = dy = - 18 x dx 8y 4 - 9x 4y - 9(2) 4(3) 3 - 18 12 2 -3. 2

=

=

=

=

Ecuacin de la Tangente: y - y1 = m (x - x1) y - 3 = - 3 (x - 2) 2 2(y - 3) = - 3 (x - 2) 2y - 6 = - 3x + 6 3x + 2y - 6 - 6 = 0 = 3x + 2y - 12 = 0 Ecuacin de la Normal:y - y1 = - 1 (x - x1) m1 y - 3 = - - 3 (x - 2) . 2 2(y - 3) = 3 (x - 2)

2y - 6 = 3x - 6 3x - 2y -6 + 6 = 0 3x - 2y = 0 Longitud de la Sub-tangente:

y1 = 3 m1 -3 2

=

- 6 =-2 3 .

.

Longitud de la Sub-normal: m1.y1 = - 3 (3) = - 9 . 2 12. 2

xy + y2 + 2 = 0 ; ( 3, - 2 ) . Derivando para obtener la pendiente en P(3, - 2 ) x.dy + y.dx + 2y.dy + 0 = 0 dx dx dx x.dy + y (1) + 2y.dy = 0 dx dx

x.dy + y + 2y.dy = 0 dx dx dy (x + 2y) = - y dx m = dy = - y dx x + 2y=

- (-2) = 2 = 2 = - 2. 3 + 2(-2) 3 - 4 - 1

Ecuacin de la Tangente: y - y1 = m (x - x1) y - (-2) = -2(x - 3) y + 2 = -2x + 6

2x + y + 2 - 6 = 0 2x + y - 4 = 0 Ecuacin de la Normal: y - y1 = - 1 (x - x1) m1 y - (-2) = - 1 (x - 3) -2 y + 2 = (x - 3 ) 2 2(y + 2) = x - 3 2y + 4 = x - 3 0 = x - 2y -3 - 4 = x - 2y - 7 = 0 Longitud de la Sub-tangente: y1 m1=

-2 -2

=

1

Longitud de la Sub-normal: m1.y1 = (-2)(-2) = 413. Clcular el rea del tringulo que forman el eje de las "x". y la tangente y la normal a la curva y = 6x - x2 en el punto (5,5).

Derivamos para encontrar la pendiente en P(5,5). y'= 6 - 2x y'= 6 - 2(5) = 6 - 10 = - 4 = m . m =- 4 . Ecuacin de la Tangente:

y - y1 = m (x - x1) y - 5 = - 4(x - 5) y - 5 = - 4x + 20 4x + y - 5 - 20 = 0 4x + y - 25 = 0Ahora encontramos la intercepcin de la tangente con el eje "x".

Cuando y = 0 ; 4x + y - 25 = 0 4x + 0 - 25 = 0 4x = 25 x = 25 . 4 N(25/4 , 0) Ecuacin de la Normal: y - y1 = - 1 (x - x1) m1 y - 5 = - 1 (x - 5) -4=

- 4(y - 5) = - (x - 5) - 4y + 20 = - x + 5

=

x - 4y + 20 - 5 = 0 x - 4y + 15 = 0Ahora encontramos la intercepcin de la Normal con el eje de las "x".

Cuando y = 0 ; x - 4y + 15 = 0 x - 4(0) + 15 = 0

x - 0 + 15 = 0 x = - 15 M( - 15, 0). Clculando la distancia MN = base del tringulo. {25 - (- 15)}2 + (0 - 0)2 4=

25 + 15 4

2

=

85 2 4

=

85 . 4

Area del Tringulo PMN = b . h . 2 Base = 85 ; h = PS = 5 485 ( 5). 4 2

=

85 . 5. 1 4(2)

=

425 unidades2. 8

14. Hallar el rea del tringulo que forman el eje de las "y" , la tangente y la normal a la curva y2 = 9 - x en el punto (5,2) Derivamos para encontrar la pendiente en el punto(5,2).y2 = 9 - x 2y.dy = 0 - dx . dx dx 2y.dy = - 1 dx m = dy = - 1 dx 2y=

-1 =-1 . 2(2) 4

Ecuacin de la Tangente: y - y1 = m(x - x1) y - 2 = - 1 (x - 5) 4

4(y - 2) = - (x - 5) 4y - 8 = - x + 5 x + 4y - 8 - 5 = 0 x + 4y - 13 = 0 El intercepto con el eje "y". Cuando x = 0 ; x + 4y - 13 = 0 0 + 4y - 13 = 0 4y = 13 y = 13 . M(0 ,13/4) 4 Ecuacin de la Normal: y - y1 = - 1 (x - x1) m1 y - 2 = - 1 (x - 5) -1 4 y - 2 = 4 (x - 5) y - 2 = 4x - 20 4x - y - 18 = 0 El intercepto de la normal con el eje "y". Cuando x = 0 ; 4x - y - 18 = 0

.

4(0) - y - 18 = 0 0 - y - 18 = 0=

- 18 = y = - 18

N(0 , - 18) Clculando la distancia MN = base del tringulo. MN = (0 - 0)2 + (- 18 - 13/4)2 Area del tringulo = b.h . 2 base = MN = 85/4. Altura = 5 ( por grfico se encontro esta altura). 85 (5) 4 2 1 85 . 5 . 1 = 425 unidades2. 4.2 8=

(- 85/4)2

=

85/4.

=

Hallar los ngulos de intercepcin de cada uno de los siguientes pares de curvas.

15.

y2 = x + 1 , x2 + y2 = 13 Primero encontramos los puntos de intercepcin. y2 = x + 1 2 2 x + y = 13 y2 = x + 1 y2 = 13 - x2

x + 1 = 13 - x2 x2 + x - 12 = 0 (x + 4) (x - 3) = 0 Interceptos: x=-4 x=3 M ( 3 , 2 ) ; R ( - 4 , 3 i ) N ( 3 , - 2 ) ; S ( - 4 , - 3 i )

Ahora encontramos las pendientes de cada curva. y2 = x + 1 2y.dy = dx + 0 dx dx 2y.dy = 1 ; dx m1 = dy = 1 . dx 2y=

Sustituyendo para (3,2). dy = 1 dx (2)(2) x2 + y2 = 13 2x + 2y.dy = 0 dx m2 = dy = - 2x dx 2y=

1 . 4

- x . Para (3,2) ; m2 = - 3 . y 2

Concluimos encontrando el ngulo de intercepcin. -3 - 1 -6 - 1 tg = m2 - m1 = 2 4 = 4 4 1 + m1 . m2 1 + ( -3 ) ( 1 ) 1 - 3 -7 4 5.

=

=

- 56 = 20

2 4 8 8 0 tg = - 14 . = arc tg ( - 14 ) = 109 39' 13". 5 5 16. y = 6 - x2 ; 7x2 + y2 = 32 Primero encontramos los puntos de intercepcin. y = 6 - x2 7x2 + y2 = 32 x2 = 6 - y x2 = 32 - y2 7 6 - y = 32 - y2 . 7 7(6 - y) = 32 - y2 42 - 7y = 32 - y2 y2 - 7y + 42 - 32 = 0 y2 - 7y + 10 = 0 (y - 5 ) (y - 2 ) = 0 y=5 y=2 Sustituyendo en y los valores de "y". x2 = 6 - y = 6 - 5 = 1 . x = 1. M(1 , 5) M(-1, 5)=

x2 = 32 - y2

32 - 4 7

=

28 7

=

4. x = 2. 7

N(2 , 2) N(-2 , 2)

Ahora encontramos las pendientes de cada curva, trabajamos

para esto con los valores positivos, M(1,5) ; N(2,2) Para N(2,2). y = 6 - x2

m2 = dy = 0 - 2x = - 2x = - 2(2) = - 4 dx 7x2 + y2 = 32 14x + 2y.dy = 0 dx m1 = dy = - 14x = - 7x dx 2y y=

- 7( 2 ) 2

=

- 7.

Concluimos encontrando el ngulo de intercepcin para (2,2). tg = m2 - m1 = - 4 - (- 7) = - 4 + 7 = 3 = 0,1034482758621 1 + m1.m2 1 + (-7)(- 4) 1 + (28) 29 = arc tg (0,1034482758621) = 50 54' 22". El valor de las pendientes de cada curva en (1,5) m1 = - 2x = - 2( 1 ) = - 2. m2 = - 7x = - 7 ( 1 ) = - 7 . y 5 5 Concluimos encontrando el ngulo de intercepcin para (1,5) Tg = m2 - m1 = - 7/5 - (- 2) 1 + m1.m2 1 + (-2)(-7/5)=

- 7/5 + 2 1 + 14/5

=

3/5 = 3 . 19/5 19

Tg = 3 = arc tg ( 3 ) = 0,1578947368421 = 80 58' 21". 19 19

17.

y = x2

; y2 - 3y = 2x

y = x2 y2 - 3y = 2x

Sustituyendo en (x2)2 - 3(x2) = 2x x4 - 3x2 - 2x = 0 x(x3 - 3x - 2 ) = 0 x=0 x3 - 3x - 2 = 0 (x + 1) (x+ 1) (x - 2) = 0 x+1=0 x=-1 x-2=0 x=2

Clcular los Mximos y Mnimos de cada una de las Funciones Siguientes. Pgina 69.1. x3 - 6x2 + 9x Primeramente derivamos: f(x) = x3 - 6x2 + 9x. f '(x) = 3x2 - 12x + 9. Luego igualamos la primera derivada igual a cero. f '(x) = 3x2 - 12x + 9 = 3(x2 - 4x + 3 ) = 0. f '(x) = (x - 3)(x - 1) = 0 de donde: x = 3 ; x = 1, estos serian los valores crticos. Luego: Para x = 3, se toma un nmero menor a 3, el ms pequeo, se sustituye en la primera derivada. x < 3 = 2,9. f '(x) = (x - 3)(x - 1)f '(x) = ( 2,9 - 3 )( 2,9 - 1 ) = (- 0,1 ) ( + 1,9) = - 0,19 .

Para esta clase de resultados, no es necesario hacer el prceso numrico, solamente interesa el signo. Asi en el caso: (- 0,1) ( + 1,9) = " - " . Este signo negativo lo almacenamos como un primer resultado. Luego:Para x = 3,se toma un nmero mayor que 3, el ms pequeo,este se sustituye en la primera derivada. x > 3 = 3,1 f '(x) = (x - 3)(x - 1). f '(x) = (3,1 - 3)(3,1 - 1) = (+) (+) = " + ". Este signo positivo seria el segundo resultado

Puesto que el signo de la derivada cambia de " - " a " + ", la funcin tiene un valor Mnimo . Para saber cuanto es este valor Mnimo, reemplaza