92548142 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 función 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 función 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 (θ) = sen 2θ + cos θ. Hallar :

a. f (0)

f (0) = sen 2 (0) + cos (0) = sen 0 + cos 0 = 0 + 1 = f (0) = 1.

b. f (1/2 π) .

f (1/2 π) = sen 2 π + cos π = sen π + cos 900= 0 + 0 = 0 . 2 2

c. f (π)

f (π) = sen 2 (π) + cos π = sen 3600 + cos 1800= 0 + (-1) = -1. f (π) = -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 .

x x2 _{+ 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 φ (z) = 4z , demostrar que: φ (z + 1) - φ (z) = 3φ (z) Primero encontramos φ (z + 1) φ (z+1) = 4z +1 Luego:φ (z + 1) - φ (z) = 4z +1 - 4z. φ (z + 1) - φ (z) = 4z.4 - 4z. φ (z + 1) - φ (z) = 4z (4 - 1) = 4z (3) = 3 (4z). Pero : φ (z) = 4z. ⇒ φ (z + 1) - φ (z) = 3φ (z).

9. Si φ (x) = ar ,demostrar que: φ (y). φ (z) = φ (y + z)

φ (y) = ay φ (z) = az

φ (y).φ (z) = ay.az= ay + z

Si: φ (x) = ax ⇒ φ (y). φ (z) = ay + z =φ (y + z).

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

z .

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

.

1+ y

Luego calculamos φ (z) , sustituyendo en φ (x) : φ (z) = log 1 - z

.

1 + z Ahora: φ (y) + φ (z) = log 1 - y + log 1 - z = log (1 - y)(1 - z) .

1 + y 1 + z (1 + y)(1 + z) φ (y) + φ (z) = log 1 - y - z + yz = (1 + yz) - (y + z) .

1+ y + z + yz (1+ yz) + (y + z)

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

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

(1 + yz) 1 + y + z 1 + yz + y + z (1 + yz) + (y + z) 1+ yz (1 + yz)

⇒ φ (y) + φ (z) = φ y + z = log (1 + yz) - (y + z) 1 + yz (1 + yz) + (y + z)

11. 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 Trigonométria : 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 - sen}2 _{h) + cos x (2 sen h. cos h)} sen x (1 - 2 sen2_{h) + cos x (2 sen h. cos h)} sen x (1 - 2 sen2_{h) + 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. sen2_{h + 2 cos x. sen h. cos h - sen x} 2 cos x. sen h. cos h - 2sen x. sen2_h

Factorando : 2 sen h (cos x. cos h - sen x. sen h)

Pero : según 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 + 3

Dividiendo númerador y denominador por x y luego sustituyendo por _∞ . 4x + 5 4 + 5 4 + 5 . lim x x = x = ∞ = 4 + 0 = 4 = 2. x→∞ 2x + 3 2 + 3 2 + 3 2 + 0 2 x x x ∞ 3. lim 4t 2 _{+ 3t + 2} = - 1 . t→0 t3 + 2t - 6 3

Se sustituye t →0 en el numerador y denominador. lim 4 (0)2 _{+ 3(0) + 2} = 0 + 0 + 2 = 2 = - 1 . t→0 (0)3 + 2(0) - 6 0 + 0 - 6 -6 3 4. lim x 2 _{h + 3xh + h}2 3 = x . h→0 2xh + 5h2 2 lim h (x 2 _{+ 3xh + h}2 ₎ = x 2 + 3xh + h2 h→0 h (2x + 5h) 2x + 5h

Se sustituye h →0 tanto en el numerador como en el denominador. lim x 2 _{+ 3x(0) + (0)}2

= x 2 + 0 + 0 = x 2 = x .x = x . h→0 2x + 5(0) 2x + 0 2x 2 x 2

5. lim 6x _{- 5x}3 2 _{+ 3} = 3

x→∞ 2x3 + 4x - 7

Primero dividimos, tanto en el numerador como en el denominador por x3_.

6 x _{- 5 x} 3 2 _{+ 3 6 - 5 + 3} . lim x 3 _x 3 _x3 = x x3 = x→∞ 2 x 3 _{+ 4 x - 7 2 + 4 - 7} x3 _x3 _x3 _x2 _x3

Luego sustituyendo x →∞ y teniendo presente que todo número para ∞= 0 .

6 - 5 + 3 . ∞ ∞ 3 = 6 - 0 + 0 = 6 = 3 . 2 + 4 - 7 2 + 0 - 0 2 ∞2 _∞3 6. lim (2z + 3k)3 _{- 4k}2 _z = 1 k →0 2z ( 2z - k )2 lim (2z) 3 _{+ 3(2z)}2 _{(3k) + 3(2z)(3k)}2 _{+ (3k)}3 _{- 4k z} 2 _. k→0 2z [(2z)2 - 2zk + (k)2] lim 8z 3 _{+ 36z}2 _{k + 54zk}2 _{+ 27k}3 _{- 4k}2 _{z .} k→0 2z (4z2 - 2zk + k2) Sustituyendo k →0 lim 8z 3 _{+ 36z}2 _{(0) + 54z (0)}2 _{+ 27 (0)}3 _{- 4(0)}2 _{z .} k→0 2z [4z2 - 2z(0) + (0)2] lim 8z 3 _{+ 0 + 0 + 0 - 0} = 8z 3 = 1 k→0 2z (4z2 - 0 + 0) 8z3 .

7. lim ax 4 _{+ bx}2 _{+ c} = 0

x →∞ dx5 + ex3 + fx

Dividiendo numerador y denominador para x4 _. a x 4 _{+ b x} 2 _{+ c a + b + c} . lim x 4 _x 4 _x4 = x2 x 4 . x →∞ en la operación. x→∞ d x 5 + e x + f3 x dx + e + f . x4 _x4 _x4 _{x x}3 a + b + c . lim ∞2 _∞ 4 = a + 0 + 0 . x→∞ d.∞ + e + f ∞ + 0 + 0 ∞ ∞3 lim a = 0 x→∞ ∞ 8. lim ax4 _{+ bx}2 _{+ c} = ∞ . x→∞ dx3 _{+ ex}2 _{+ fx + g}

Dividiendo numerador y denominador para x4_. a x 4 _{+ b x} 2 _{+ c a + b + c} . lim x 4 _x 4 _x4 = x x2 4 . x→∞ d x 3 _{+ e x} 2 _{+ f x + g d + e + f + g} . x4 _x4 _x4 _x4 _{x x}2 _x3 _x4 Sustituyendo x →∞ en la operación. lim a + b + c . x→∞ ∞ 2 _∞ 4 = a + 0 + 0 = a =∞ d + e + f + g 0 + 0 + 0 + 0 0 ∞ ∞2 _∞3 _∞4

9. lim s 4 _{- a} 4 = 2a2 s→a s2 - a2 lim (s _{+ a}2 2 _{) (s} 2 _{- a} 2 ₎ = s2 + a2. s→a ( s2 - a2 ) Sustituyendo s →a en la operación. lim a2 _{+ a}2 = 2a2 s→a 10. lim x 2 _{+ x - 6} = 5 . x→2 x2 - 4 4 lim (x + 3) (x - 2) = (x + 3) . Sustituyendo x→2 : x→2 (x + 2) (x - 2) (x +2) lim (2 + 3) = 5 . x→2 (2 + 2) 4 11. lim 4y2 _{- 3} = 0 y→∞ 2y3 _{+ 3y}2 Dividimos para y3_. 4 y 2 _{- 3 4 - 3.} lim y 3 _y3 = y y3 . y→∞ 2 y 3 + 3 y 2 2 + 3 y 3 _y 3 _y Sustituyendo y→∞ en la operación : 4 - 3 . lim ∞ ∞ = 0 - 0 = 0 = 0 y→∞ 2 + 3 2 + 0 2

∞

12. lim 3h + 2xh2 _{+ x} 2 _h 3

= - 1 . h→∞ 4 - 3xh - 2x3h3 2x Dividiendo todo para h3_.

3h + 2xh 2 _{+ x} 2 _h 3 _{3 + 2x + x}2 lim h _h3 3 _h3 = h 2 h . h→∞ 4 - 3xh - 2x 3 _h 3 _{4 - 3x - 2x}3 h3 _h3 _h3 _h3 _h2 Sustituyendo h→∞ en la operación : lim 3 + 2x + x2 _{3 + 2x + x}2 h→∞ ∞ 2 _∞ = ∞ ∞ = 0 + 0 + x2 = x 2 = - 1 . 4 - 3x - 2x3 _{4 - 3x - 2x}3 _{0 - 0 - 2x}3 _-2x3 _2x ∞3 _∞2 _{∞ ∞} 13. lim aox n + a 1x n-1 + … + a n = ao . x→∞ boxn + b1xn-1 + … + bn bo

lim aox n + a 1x n .x + … + a-1 n . Dividiendo todo para el mayor exponente xn x→∞ boxn + b1xn.x-1 + … + b aox n + a1x n .x -1 + … + an ao + a1.x-1 + … + an . lim x n x n xn = xn . x→∞ box n + b1.x .xn + … + b-1 n bo + b1.x-1 + … + bn xn _xn _xn _xn Sustituyendo ∞ en x. ao + a1 + … + an lim ∞ ∞ = lim ao + 0 + … + 0 = ao . x→∞ bo + b1 + … + bn x→∞ bo + 0 + … + 0 bo ∞ ∞

14. lim aox + an 1x n-1 + … + an = an x→0 boxn + b1xn-1 + … + bn bn Sustituyendo x→0 en x lim ao ( 0 )n + a 1 ( 0 )n-1 + … + a n = ao (0) + a1 (0) + … + an . x→0 b0 ( 0 )n + b1 ( 0 )n-1 + … + b n bo (0) + b1 (0) + … + bn lim 0 + 0 + … + an = an . x→∞ 0 + 0 + … + bn bn 15. lim (x + h)n _{- x} n = nxn-1 h→0 h

Desarrollando el Binomio de Newton.

lim xn _{+ nx}n-1_{h + n(n-1).x}n-2 _{.h + n(n-1)(n-2).x}2 n-3 _.h 3 _{+ … + h}n _{- x}n_.

h→0 1x2 1x2x3

lim nxn-1_{.h + n(n-1).x}n-2_.h2 _{+ n(n-1)(n-2).x}n-3_.h3 _{+ … + h}n h→0 2 6

Dividiendo todo para h .

lim nx n-1 _{. h + n(n-1).x}n-2 _{. h + n(n-1)(n-2).x}2 n-3 _{. h + … + h} 3 _.n h→0 h 2 h 6 h h . lim nxn-1 _{+ n(n-1).x}n-2 _{.h + n(n-1)(n-2).x}n-3 _.h 2 _{+ … + h}n-1 h→0 2 6 Sustituyendo h→0 en la operación. lim nxn-1 _{+ n(n-1).x}n-2 _{( 0 ) + n(n-1)(n-2).x}n-3 _{( 0 )}2 _{+ … + ( 0 )}n-1 h→0 2 6 lim nxn-1 _{+ 0 + 0 + … + 0} = nxn-1 h→0

16. lim √ x + h - √ x = 1 . h→0 h 2 √x Racionalizando el numerador: lim ( √ x + h - √ x ) ( √ x + h + √ x ) h→0 h (√x + h + √x) lim ( √ x + h ) _{- (}2 _{√ x )} 2 = x + h - x . h→0 h(√x + h + √x) h(√x+h + √x) lim h = 1 . h→0h (√x + h + √x) (√x + h + √x ) Sustituyendo h→0 en la operación. lim 1 = 1 = 1 . h→0 (√x + 0 + √x ) (√x + √x ) 2√x

17. Dado f (x) = x2 , demostrar que :

lim f (x+h) - f (x) = 2x h→0 h Si f (x) = x2 f (x+h) = (x+h)2 ⇒ lim (x + h)2 _{- x} 2 = x + 2xh + h2 - x2 2 = 2xh + h2 = h (2x + h) = 2x + h h→0 h h h h . Sustituyendo h →0en la operación: lim 2x + h = 2x + 0 = 2x h→0

18. Dado f (x) = ax2 + bx + c , demostrar que: lim f (x + h) - f (x) = 2ax + b. h→0 h 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 función:

lim f (x + h) - f (x) = ax 2 + 2axh + ah2 + bx + bh + c - (ax + bx + c)2 .

h→0 h h

lim ax 2 _{+ 2axh + ah}2 _{+ bx + bh + c - ax - bx - c}2

= 2axh + ah2 + bh .

h→0 h h

lim h (2ax + ah + b ) = 2ax + ah + b ;

h→0 h . lim 2ax + a ( 0 ) + b = 2ax + b

h→0

19. Dado f (x) = 1 ,demostrar que :

x lim f (x + h) - f (x) = - 1 . h→0 h x2 f(x) = 1 . x f(x+h) = 1 . x + h

1 - 1 x - (x + h) x - x - h - h . lim x + h x = x (x + h) = x (x + h) = x (x + h) = h→0 h h h h . 1 1 1 lim - 1 . h→0 x (x + h).

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

lim f (x+h)3 _{- f (x)} = x + 3x3 2 h + 3xh2 + h 3 - x 3 = 3x 2 h + 3xh2 + h 3 . h→0 h h h lim h (3x 2 _{+ 3xh + h}2 ₎ = 3x2 + 3xh + h2 h→0 h . Sustituyendo h→0 en la operación. lim 3x2 _{+ 3x ( 0 ) + ( 0 )}2 = 3x2 + 0 + 0 = 3x2 h→0

Problemas “ Página 32 “

Calcular la derivada de cada una de las siguientes funciones usando la regla general.

1. y = 2 - 3x

Se sustituye en la función "x" por "x + ∆x" y se calcula el nuevo valor de la función y + ∆y .

y + ∆y = 2 - 3 (x + ∆X) . Se resta el valor dado de la función del

nuevo valor y se obtiene ∆y. Y + ∆y = 2 - 3 (x + ∆X) Y + ∆y = 2 – 3X -3∆X y + ∆y - y =2 - 3x - 3∆x – 2 + 3X . ∆y = - 3∆x. Se divide ∆y para ∆x. ∆ y = - 3 ∆ x ∆x ∆x

Se calcula el límite de este cociente cuando ∆x → 0 . El límite así hallado es la derivada buscada.

∆ y = - 3 ∆ x . ∆X ∆x . lim ∆x→0 dy = - 3 Y’ = -3 dx

2. y = mx + b. y + ∆y = m (x + ∆x) + b. Y + ∆y = mX + m∆x + b y + ∆y - y=mx + m∆x + b – mX - b ∆y = m∆X ∆ y = m ∆ x ∆x ∆x ∆ y = m ∆x lim ∆x→0 dy = m . Y ‘ = m dx 3. y = ax2 y + ∆y = a ( X + ∆x)2. y + ∆y = a ( X2 +2X ∆X + ∆x2 ) y + ∆y = aX2 +2aX ∆X + a∆x2 y + ∆y - y=aX2+2aX ∆X + a∆x2- aX2 ∆y = 2aX ∆X + a∆x2 ∆ y = 2ax. ∆ x + a . ∆ x 2 . ∆x ∆x ∆x . ∆ y = 2ax + a.∆x . ∆x lim ∆x→0 dy = 2ax + a (0) dx dy = 2ax . Y ‘ = 2 ax dx

4. s = 2t - t2. s + ∆s = 2(t + ∆t) - (t + ∆t)2. s + ∆s = 2t + 2∆t - (t2 +2t ∆t + ∆t 2 ) s + ∆s = 2t + 2∆t - t2 -2t ∆t - ∆t 2 s + ∆s - s = 2t + 2∆t - t2 _{-2t ∆t - ∆t} 2 _{– 2t +t}2 ∆s = 2. ∆t - 2t. ∆t - ∆t 2 ∆ s = ∆ t (2 - 2t - ∆ t) ∆t ∆t . ∆ s = 2 - 2t - ∆t ∆t lim ∆x→0 ds = 2 - 2t - 0 Y’ = 2 – 2t dt 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= 3cx . 2 x + 3cx. ∆ ∆ x + c 2 ∆ x 3 ∆x ∆x ∆x ∆x . ∆ y = 3cx2 + 3cx.∆x + c∆x2. ∆x lim ∆x→0 dy = 3cx2 + 3cx( 0 ) + c ( 0 )2 dx Y ‘ _{= 3cx}2

6. y = 3x - x3. y + ∆y = 3 (x + ∆x) - (x + ∆x)3. y + ∆y = 3 x + 3∆x - (x3 +3X2∆x + 3X ∆x2 + ∆x3 ) y + ∆y = 3 x + 3∆x - x3 -3X2∆x - 3X ∆x2 - ∆x3 y + ∆y - y =3 x + 3∆x - x3 -3X2∆x - 3X ∆x2 - ∆x3– 3X + X3. ∆y =3∆x - 3X2∆x - 3X ∆x2 - ∆x3 ∆ y= 3.∆ x - 3x 2 . x - 3x.(∆ ∆ x) - (2 ∆ x) 3 . ∆x ∆x ∆x ∆x ∆x ∆ y = 3 - 3x2 - 3x (0) - (∆x)2 ∆x lim ∆x→0 dy = 3 - 3x2 . Y’ = 3 – 3x2 dx 7. u = 4v2 + 2v3. u + ∆u = 4 (v + ∆v)2 + 2 (v + ∆v)3. u + ∆u = 4 ( v2 _{+ 2v∆v + ∆v}2 _{) + 2 ( v}3 _{+ 3v}2_{∆v + 3v∆v}2 _{+ ∆v}3 ₎ u + ∆u = 4 v2 _{+ 8v∆v +4∆v}2 _{+ 2v}3 _{+ 6v}2_{∆v + 6v∆v}2 _{+ 2∆v}3 u + ∆u – u = 4 v2 _{+ 8v∆v +4∆v}2 _{+ 2v}3 _{+ 6v}2_{∆v + 6v∆v}2 _{+ 2∆v}3 _{– 4v}2 _{– 2v}3 ∆u = 8v∆v +4∆v2 + 6v2∆v + 6v∆v2 + 2∆v3 ∆ u= 8v. ∆ v + 4. ∆ v 2 + 6v . 2 ∆v + 6v. ∆ v 2 + 2. ∆v 3 . ∆v ∆v ∆v ∆v ∆v ∆v ∆ u = 8v + 4. ∆v + 6v2 + 6v. ∆v + 2. ∆v2 ∆v ∆ u= 8v + 4(0) + 6v2 + 6v(0) + 2(0 )2 ∆v lim ∆v→0 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 + 4X3∆x + 6X2∆x2 + 4X∆x3 + ∆x4 y + ∆y - y = x4 + 4X3∆x + 6X2∆x2 + 4X∆x3 + ∆x4 - x4 ∆y = 4x3. ∆x + 6x2.∆x2 + 4x.∆x3 + ∆x4. ∆ y = 4x 3 . ∆ x + 6x 2 ( x)∆ 2 + 4x ( ∆ x) 3 + ( ∆ x) 4 ∆x ∆x ∆x ∆x ∆x . ∆ y= 4x3 + 6x2. ∆x + 4x (∆x)2 + (∆x)3 ∆x ∆ y = 4x3 + 6x2(0) + 4x(0)2 + ( 0 )3 ∆x lim ∆x→0 dy = 4x3 . Y ‘ = 4x3 dx 9. e = 2 . θ + 1 e + ∆e = 2 . (θ + ∆θ) + 1 e + ∆e - e = 2 - 2 . (θ + ∆θ) + 1 (θ+1) ∆e = 2 ( θ + 1) - 2[(θ + ∆θ ) + 1] . [(θ + ∆θ) + 1] (θ + 1) ∆e = 2 θ + 2 - 2θ - 2. ∆θ - 2 = - 2. ∆θ . [(θ + ∆θ) + 1] (θ + 1) [(θ + ∆θ) + 1](θ + 1) ∆ e = - 2.∆θ = -2 =

∆θ [(θ + ∆θ) + 1](θ + 1)(∆θ) [(θ + ∆θ) + 1](θ + 1) ∆ e = - 2 . ∆θ [(θ + 0) + 1](θ + 1) lim ∆x→0 de = -2 = dθ (θ + 1) (θ + 1) de = -2 . dθ (θ + 1)2 Y ‘ = - 2 . (θ + 1)2 10. y = 3 . x2 _{+ 2} y + ∆y = 3 . (x + ∆x)2 _{+ 2} y + ∆y - y= 3 - 3 . (x + ∆x)2 _{+ 2 x}2 _{+ 2} ∆y = 3 (x2 + 2) - 3 [(x + ∆ x) 2 + 2] [(x + ∆x)2 _{+ 2] (x}2 _{+ 2)} ∆y = 3x 2 + 6 -3 [x + 2x. 2 ∆ x +(∆ x) 2 +2] [(x + ∆x)2 _{+ 2] (x}2 _{+ 2)} ∆y =3x 2 + 6 - 3x 2 - 6x. ∆x - 3(∆ x) 2 - 6 = - 6x. ∆ x - 3(∆ x) 2 . [(x + ∆x)2 _{+ 2] (x}2 _{+ 2) [(x+∆x)}2 _{+ 2] (x}2 _{+ 2)} ∆ y = ∆ (- 6x -3. x ∆ x) = - 6x - 3. ∆ x = ∆x [(x + ∆x)2 _{+ 2] (x}2 _{+ 2) ∆x [(x + ∆x)}2 _{+ 2] (x}2 _{+ 2)}

∆ y = - 6x - 3 (0) . ∆x [(x + 0)2 _{+ 2] (x}2 _{+ 2)} lim ∆x→0 dy = - 6x - 0 = - 6x . dx (x2 _{+ 2) (x}2 _{+ 2) (x}2 _{+ 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 = t 2 + t. t + 4t -(t∆ 2 + 4t + t. ∆ t + 4. t)∆ = (t + ∆t) t ∆s = t 2 + t. ∆ t + 4t - t 2 - 4t - t. ∆t - 4. ∆ t = - 4. ∆ t . (t + ∆t) t (t + ∆t) t ∆ s = - 4 ( ∆ t ) . ∆t (t + ∆t) t ( ∆t ) ∆ s = - 4 . ∆t (t + 0)t lim ∆t→0

ds = - 4 = - 4 . dt t.t t2 S ´= -4 / t2 12. y = 1 . 1 - 2x y + ∆y = 1 . 1 - 2(x + ∆x) y + ∆y - y= 1 - 1 . 1 - 2 (x + ∆x) 1 - 2x ∆y = (1 - 2x) - [1 -2(x + ∆ x)] = 1 - 2x -(1 - 2x - 2∆ x) = [1 - 2(x+∆x)](1 - 2x) [1 - 2(x+∆x)](1 - 2x) ∆y = 1 - 2x - 1 + 2x + 2 ∆ x = 2∆ x . [1 - 2(x+∆x)](1 - 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) ∆ y = 2 . ∆x [1 - 2 (x + 0)](1 - 2x) lim ∆x→0 dy = 2 = 2 . dx (1 - 2x) (1 - 2x) (1 - 2x)2

dy = 2 dx (1 - 2x)2 13. e = θ . θ + 2 e + ∆e = θ + ∆θ . (θ + ∆θ) + 2 e + ∆e - e= θ + ∆θ _ θ . (θ + ∆θ) + 2 θ + 2 ∆e =( + 2) ( θ + θ ∆θ ) - [(θ + θ ∆θ ) + 2] = θ 2 + 2 θ + θ . ∆θ + 2∆θ - θ 2 - θ . ∆θ - 2 θ . [(θ + ∆θ) + 2](θ + 2) [(θ + ∆θ) + 2] (θ + 2) ∆e = 2 ∆θ . [(θ + ∆θ) + 2] (θ + 2)

Dividiendo a ambos miembros para ∆θ y simplificando : ∆ e = 2∆θ = 2. ∆θ . ∆θ [(θ + ∆θ) + 2] (θ + 2). ∆θ [(θ + ∆θ) + 2] (θ + 2). ∆θ . ∆ e= 2 = 2 = 2 . ∆θ [(θ + 0) + 2] (θ + 2) (θ + 2) (θ + 2) (θ + 2)2 lim ∆θ →0 de = 2 = 2 . dθ (θ + 2) (θ + 2) (θ + 2)2 14. s = At + B Ct + D

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 + AD ∆t + BCt + BD . [C(t + ∆t) + D] (Ct + D)

- ACt2 _{- BCt - ACt ∆t - BC∆ t - Adt - BD .}

[C(t + ∆t) + D] (Ct + D) ∆s = A.D. ∆ t - B.C. ∆ t = ∆ t (A.D - B.C) . [C(t + ∆t) + D] (Ct + D) [C(t + ∆t) + D] (Ct + D) ∆ s = ∆ t (A.D - B.C) = (A.D - B.C) . ∆t [C(t + ∆t) + D] (Ct + D) (∆t) [C(t + ∆t) + D] (Ct + D) ∆ s = (A.D - B.C) = (A.D - B.C) . ∆t [C(t + 0) + D] (Ct + D) (Ct + D)(Ct + D) lim ∆t→0 ds = (A.D - B.C) = (A.D - B.C) . dt (Ct + D)(Ct + D) (Ct + D)2 15. y = x 3 + 1 . x

y + ∆y = (x + ∆ x) 3 + 1 . (x + ∆x) y + ∆y - y = (x + ∆ x) 3 + 1 - x 3 + 1 (x + ∆x) x ∆y = [(x + ∆ x) 3 + 1] x - (x3 + 1) (x + ∆ x) (x + ∆x) x ∆y = {[x 3 + 3x2 ∆ x + 3x(∆ x) + (2 x)∆ 3 ] + 1}(x) - x - x4 3 ( x) - x - ∆ ∆ x (x + ∆x) x ∆y = x 4 + 3x 3 ∆ + 3xx 2 ( ∆ x) 2 + ( x)∆ 3 (x) + x - x 4 - x 3 ( ∆ - x) x - ∆ x (x + ∆x) x ∆y = 2x 3 . x + 3x∆ 2 ( ∆ 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 [2x 3 + 3x 2. ∆ x + ( ∆ x) 2 .x - 3x2 ∆. x + ( ∆ x) 2 .x - 1] ∆x (x + ∆x)(x)( ∆x) ∆ y= [2x + 3x3 2 ( 0 ) + (0)2 (x) - 3x2 (0) + (0)2 (x) - 1] ∆x ( x + 0 ) x lim ∆x→0 y’ = 2x 3 + 0 + 0 - 0 + 0 - 1 = 2x 3 - 1 = 2 x 3 - 1 x . x x2 x2 x2

dy = 2x - 1 dx x2 16. y = 1 . x2 _{+ a}2 y + ∆y = 1 . (x + ∆x)2 _{+ a}2 y + ∆y - y = 1 _ 1 . (x + ∆x)2 _{+ a}2 _x2 _{+ a}2 ∆y = 1 (x 2 + a ) - 1 [(x + 2 ∆ x) 2 + a 2 ] = x 2 + a 2 -[x 2 + 2x.∆ x + (∆ x) 2 + a ]2 [(x + ∆x)2 _{+ a}2_{] (x}2 _{+ a}2_{) [(x+∆x)}2 _{+ a}2_{] (x}2 _{+ a}2₎ ∆y =x 2 + a 2 - x 2 - 2x. ∆ x -( ∆ x)2 - a 2 = - 2x. ∆ x -( ∆ x) 2 . [(x + ∆x)2 _{+ a}2_](x2 _{+ a}2_{) [(x + ∆x)}2 _{+ a}2_{] (x}2 _{+ a}2₎ ∆y = - ∆ x (2x + ∆ x) . [(x + ∆x)2 _{+ a}2_{] (x}2 _{+ a}2₎

Dividiendo a ambos miembros para ∆x, tenemos :

∆ y= - ∆ x (2x + ∆ x) = - (2x + ∆ x) . ∆x [(x + ∆x)2 _{+ a}2_](x2 _{+ a}2_{). ∆x [(x + ∆x)}2 _{+ a}2_{] (x}2 _{+ a}2₎ ∆ y = - (2x + 0) . ∆x [(x + 0)2 + a2] (x2 + a2) lim ∆x→0 dy = - 2x = - 2x . dx (x2 _{+ a}2_{) (x}2 _{+ a}2_{) (x}2 _{+ a}2 ₎2 dy = - 2x dx (x2 _{+ a}2 ₎2

17. y = x . x2 _{+ 1} y + ∆y = x + ∆ x . (x + ∆x)2 _{+ 1} y + ∆y - y = x + ∆ x - x . [(x + ∆x)2 _{+ 1] (x}2 _{+ 1)} ∆y = (x + ∆ x) (x2 + 1) - x [(x + ∆ x) 2 + 1] . [(x + ∆x)2 _{+ 1] (x}2 _{+ 1)} ∆y = x 3 + x + ∆ x. x 2 + ∆ x - x [x2 + 2x. ∆ x + (∆ x) 2 + 1] [(x + ∆x)2 _{+ 1] (x}2 _{+ 1)} ∆y = x 3 + x + ∆ x. x 2 + ∆ x - x 3 - 2. ∆ x. x 2 - x. ( ∆x) 2 - x . [(x + ∆x)2 _{+ 1] (x}2 _{+ 1)} ∆y = - ∆ x.x 2 - x.(∆ x) 2 + ∆ x = - ∆ x (x 2 + x . ∆ x - 1 ) . [(x + ∆x)2 _{+ 1] (x}2 _{+ 1) [(x + ∆x)}2 _{+ 1](x}2 _{+ 1)} ∆ y= - ∆ (xx 2 + x. ∆ x - 1) = - (x2 + x. ∆ x - 1) . ∆x [(x + ∆x)2 _{+ 1](x}2 _{+ 1) . ∆x [(x + ∆x)}2 _{+ 1](x}2 _{+ 1)} ∆ y = - [x2 + x(0) - 1] . ∆x [(x + 0)2 + 1] (x2 + 1) lim ∆x→0 dy = -(x - 1) 2 = 1 - x2 dx (x2 _{+ 1) (x}2 _{+ 1) (x}2 _{+ 1)}2

18. y = x 2 . 4 - x2 y + ∆y = (x + ∆ x) 2 . 4 - (x + ∆x)2 y + ∆y - y = (x + ∆ x) 2 - x2 . [4 - (x + ∆x)2_{] (4 - x}2₎ ∆y = (x + ∆ x) 2 (4 - x ) - [4 - (x + 2 x)∆ 2 ] x 2 . [4 - (x + ∆x)2_{] (4 - x}2₎ ∆y = [x 2 + 2x. ∆ x + ( x)∆ ](4 - x2 2 ) - [4-(x2 +2x. ∆ x + (∆ x) 2 ]( x 2 ) [4 - (x + ∆x)2_{] (4 - x}2₎ ∆y = 4x 2 + 8x. ∆ x + 4(∆ x) 2 - x 4 - 2x . 3 ∆ x - x2 .( x)∆ 2 -[4 -x2 -2x. x-(∆ ∆ x) 2 ](x ) 2 [4 - (x + ∆x)2_{] (4 - x}2₎ ∆y = 4x 2 + 8x. ∆ x + 4( ∆x) 2 x 4 - 2x . 3∆ x - x . ( 2 ∆ x) 2 - 4x 2 + x 4 + 2x 3 . ∆ x + x 2 . ( x)∆ 2 [4 - (x + ∆x)2_{](4 - x}2₎ ∆y = 8x.∆ x + 4. (∆ x) 2 = ∆ x (8x + 4. x) ∆ . [4 - (x + ∆x)2_{](4 - x}2_{) [4 - (x + ∆x)}2_{](4 - x}2₎

Dividiendo, para ∆x , tenemos : ∆ y= ∆ x (8x + 4. ∆ x) . ∆y [4 - (x + ∆x)2_{](4 - x}2_{) . ∆x} . ∆ y = 8x + 4. ∆ x = 8x + 4( 0 ) = ∆x [4 - (x + ∆x)2](4 - x2) [4 - (x + 0)2](4 - x2) lim ∆x→0

dy = 8x + 0 = 8x dx [4 - x2_{] (4 - x}2 _{) (4 - x}2 ₎2 19. y = 3x2 - 4x - 5. y + ∆y = 3 (x + ∆x)2 - 4 (x + ∆x) - 5 y + ∆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)[6x + 3 ( x) - 4]∆ =∆ x [6x + 3 ( ∆ x) - 4]= 6x + 3(0) - 4 ∆x ∆x ∆x lim ∆x→0 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 = a (t + ∆t)2 + b (t + ∆t) + c - (at2 + bt + c) . ∆s = 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 ∆t ∆ s = 2at + a( 0 ) + b = 2at + 0 + b . ∆t lim ∆t→0 ds = 2at + b . dt 21. u = 2v3 - 3v2 u + ∆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 [6v 2 + 6v. ∆ v + 2. ( ∆ v) 2 - 6v - 3. ∆ v] ∆v ∆v . ∆ u = 6v2 + 6v. ∆v + 2 (∆v)2 - 6v - 3. ∆v ∆v lim ∆v→0

∆ 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 + 3x2∆x + 3x(∆x)2 + (∆x)3] + b[x2 + 2x. ∆x + (∆x)2 + cx + c. ∆x + d - (ax3 _{+ bx}2 _{+ cx + d)}

∆y = ax3+ 3ax2.∆x + 3ax.(∆x2) + a.(∆x)3+ bx2 + 2bx.∆x + b(∆x)2

+ cx + c. ∆x + d - ax3 _{- bx}2 _{- 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 ∆x

∆ y = 3ax2 + 3ax ( 0 ) + a.( 0 )2 + 2bx + b.( 0 ).∆x + c ∆x lim ∆v →0 dy = 3ax2 + 0 + 0 + 2bx + 0 + c dx 23. e = (a - bθ)2 e + ∆e = [a - b (θ + ∆θ )]2 e + ∆e - e = [a - b (θ + ∆θ )]2 - (a - bθ)2

∆e = (a - bθ - b.∆θ)2 - (a - bθ)2 ∆e = [a + (-bθ) + (- b.∆θ)]2 - (a - bθ)2

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

∆e = a2 + (bθ)2+(b.∆θ)2- 2a(bθ) -2a(b.∆θ) +2(bθ)(b.∆θ) - a2 + 2a.bθ - (bθ)2

∆e = (b. ∆θ)2 -2a(b. ∆θ) + 2(bθ)(b. ∆θ) ∆e = b2(∆θ)2 - 2a(b. ∆θ) + 2(bθ)(b.∆θ) Factorando y dividiendo para ∆θ. ∆ e = ∆θ {b 2 .( ∆θ ) - 2a.b + 2b2 . θ } ∆θ ∆θ . . ∆ e = b2.(∆θ) - 2a.b + 2b2.θ= b2.(0) - 2a.b + 2b2.θ ∆θ lim ∆θ →0 de = 0 - 2ab + 2b2.θ = 2b2.θ - 2ab = 2b (b.θ - a ) dθ 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 = - 5 + 4x + 2( 0 ) ∆x lim ∆θ →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 = [A (x + ∆x) + B] [C (x + ∆x) + D] - (Ax + B) (Cx + D). y - y + ∆y = (Ax + A. ∆x + B ) (Cx + C. ∆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 ∆x .

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

lim ∆x →0

∆ y = 2ACx + 0 + AD + BC

∆θ

lim ∆x→0 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 + b∆t]3 - [a3 + 3a2bt + 3a(bt)2 +(bt)3] ∆s =a3 + (bt) 3 + (b∆t)3+ 3a2(bt) + 3a2(b∆t) + 3a(bt) 2+ 3(bt)2(b∆t) +

3a(b∆t)2 _{+ 3(bt)(b}∆_t)2 _{+ 6a(bt) (b∆t) - a} 3_{- 3a}2_{(bt) - 3a(bt)} 2 _{- (bt)} 3

∆s = (b∆t)3+ 3a2(b∆t) + 3(bt)2 (b∆t) + 3a (b∆t)2 + 3 (bt) (b∆t)2 + 6a (bt) (b∆t)

Factorando , dividiendo y simplificando para ∆t .

∆ s= ∆ t ∆ {(b t) 2 + 3a2 b + 3b3 t2 + 3ab2 ∆ t + 3b3 t. ∆ t + 6ab2 t} ∆t ∆t . ∆ s = [b (0)2 + 3a2b + 3b3t2 + 3ab2.(0) + 3b3t.(0) + 6ab2t ∆t lim ∆t→0 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 . a + b (x + ∆x)2 y + ∆y - y = x + ∆ x - x . a + b (x + ∆x)2 _{a + bx}2 ∆y = (x + ∆ x) (a + bx2 ) - x {a + b (x + x)∆ 2 } [a + b (x + ∆x)2_{] [a + bx}2_] ∆y = ax + bx3 + a. x + bx∆ 2 . x - x{a + b[x∆ 2 + 2x.∆ x + (∆ x) 2 ]} [a + b (x + ∆x)2_{] [a + bx}2_] ∆y = ax + bx3 + a. ∆ x + bx2 . x - x{a + bx∆ 2 + 2bx.∆ x + b.( x)∆ 2 } [a + b (x + ∆x)2_{] [a + bx}2_] ∆y =ax + bx 3 + a.∆ x + bx . 2 x ∆ - ax - bx 3 - 2bx . 2 x ∆ - bx.(∆ x) 2 . [a + b (x + ∆x)2_{] [a + bx}2_]

∆y = a. x - bx∆ . 2∆ x - bx.(∆ x) . Factorando y dividiendo para 2 ∆x:

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

∆x [a + bx2] [a + bx2] ∆x→0 dy = a - bx 2 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 } (x 2 ) - (x + ∆ x) 2 (a + bx2 ) (x + ∆x)2 _x2 ∆y ={a + b[x2 +2x.∆ x + (∆ x) ]}(x2 2 ) - {x + 2x.2 ∆ x + (∆ x) 2 }(a +bx2 ) (x + ∆x)2 _x2 ∆y ={a + bx2 + 2bx . ∆ x + b. (∆ x) 2 }(x 2 ) - {ax2 + bx4 + 2ax. ∆ x (x + ∆x)2 _x2 _{+ 2bx} 3 _{. ∆ x + a (∆ x)} 2 _{+ bx}2 _{.( ∆ x)} 2 _} (x + ∆x)2 _x2 ∆y =ax + 2 bx 4 + 2bx 3 . ∆ x + bx 2( ∆x) 2 ax - 2 - bx 4 -2ax.∆ x - (x + ∆x)2 _x2 ∆y = 2bx 3 . x - a(∆ ∆ x) 2 - bx 2 . ( ∆ x) 2 (x + ∆x)2 _x2 ∆y = - 2ax.∆ x - a( x)∆ 2 (x + ∆x)2 _x2

Factorando , dividiendo y simplificando para ∆x : ∆ y=∆ x {-2ax - a ( ∆ x)} = (∆ x) {-2ax - a ( ∆ x)} ∆x (x + ∆x)2_{. x}2_{. (∆x) (x + ∆x)}2_{. x}2_{. (∆x)} ∆ y = {-2ax - a (∆ x)} = - 2ax - a ( 0 ) = - 2ax - 0 ∆x (x + ∆x)2. x2 (x + 0 )2 .x2 x2.x2 lim ∆x→0

dy = - 2ax = - 2a. x = - 2a dx x 4 _x3_{.x x}3 29. y = x2 . a + bx2 y + ∆y = (x + ∆ x) 2 . a + b (x + ∆x)2 y + ∆y - y = (x + ∆ x) 2 - x2 . [a + b (x + ∆x)2_{] (a + bx}2₎ ∆y = (x + ∆ x) 2 (a + bx ) - {a + b (x + 2 ∆ x) 2 }( x ) 2 [a + b (x + ∆x)2 _{] ( a + bx}2 ₎ ∆y = {x 2 + 2x. ∆ x + (∆ x) }(a + bx2 2 ) - {a + b [x2 + 2x. ∆ x + (∆ x) 2 ]}( x2 ) [a + b (x + ∆x)2 _{]( a + bx}2 ₎

∆y = {ax 2 +bx +2ax(4 x)+2bx∆ 3 ( ∆ x)+a(∆ x) +bx2 ( 2∆ x) 2 }-{a+bx +2bx(2 ∆ x)+b(∆ x) 2 }(x 2 )

[a + b (x + ∆x)2 _{] ( a + bx}2₎

∆y = ax 2 + bx 4 +2ax. x+∆ 2bx 3 ∆ x +a( ∆ x) 2 + bx ( 2∆ x) 2 - ax 2 + bx +4 2bx 3 . ∆ x + bx 2 ( x)∆ 2 [a + b (x + ∆x)2 _{] ( a + bx}2 ₎

∆y = 2ax.∆ x + a(∆ x) 2 .

[a + b (x + ∆x)2 _{] ( a + bx}2 ₎

Factorando , dividiendo y simplificando para ∆x: ∆y = (∆ x) {2ax + a( x)} ∆ = [a + b (x + ∆x)2_{]( a + bx}2 _)(∆x) ∆y = ( ∆ x) {2ax + a(∆ x)} . [a + b (x + ∆x)2_{](a + bx}2_{) (∆x)} . ∆ y = (2ax + a. x) ∆ = 2ax + a ( 0 ) . ∆x [a + b (x + ∆x)2 _{] ( a + bx}2_{) [a + b (x + 0)}2_{] (a + bx}2₎

lim ∆x→0

dy = 2ax + 0 = 2ax dx (a + bx2_{) (a + bx}2_{) (a + bx}2₎2

Problemas - Paginas : 34 y 35

Aplicando las Derivadas, hallar la pendiente y la inclinación 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 α= 2 = m. α = 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 α= - 1 α = 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 α= - 4 α = arc tg (- 4) α = 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 . α = 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 α= - 3 .

α = arc tg ( - 3 ) = 108o 26' 5''

6. Hallar el punto de la curva y=5x - x2 en el que la inclinación

de la tangente es de 45o_.

y = 5x - x2. Según dato del problema tg 45º= 1 . y' = 5 - 2x. ⇒ m = 1 .

m = y' = 5 - 2x = 1.

solucionando la ecuación: 5 - 2x = 1 ; 5 - 1 = 2x 2 = x ; x = 2 .

Sustituyendo x = 2 en la ecuación 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 = 4x. y' = 3x2 + 1. y' = 4

m1 = 3x2 + 1. 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 . y = x3 + x y1= (1)3 + (1) y2= (-1)3 + (-1) y1= 1 + 1 y2= -1 -1 = -2 y1= 2 y2 = -2 ⇒ P1 (1 , 2) ⇒ P2 (-1 , -2)

En cada uno de los siguientes problemas hallar:

a) Los puntos de intercepción del par de curvas dado.

b) La pendiente y la inclinación de la tangente a cada curva, y el ángulo formado por las tangentes en cada punto de intercepción. 8. 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 m2= 2x m1 = - 2(1) = - 2 m2= 2(1) =

m1 = - 2 m2= 2 tg θ= m1 - m2 = - 2 - 2 = - 4 = - 4 = 4 . 1+ m1.m2 1 + (-2) (2) 1 - 4 - 3 3 tg θ= 4 ; θ = arc tg 4 = 53º 8' 3 3 Cuando: x = -1 m1= - 2x = - 2(-1) = 2 m2= 2x = 2 (-1) = -2 m1= 2 ; m2= - 2 tg θ = m1 - m2 = 2 - (-2) = 2 + 2 = 4 = - 4 . 1+ m1.m2 1 + (2) (-2) 1 - 4 - 3 3 θ= arc tg (- 4/3) θ = 126º52' 11" Puntos de intercepción 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 P2 = ( -1 , 0 ) y = 1 - 1 y = 0

9. y = x2 (1)

x - y + 2 = 0 (2)

Igualamos las 2 curvas en función de ''y'' para encontrar sus intercepciones. y = x2. (1) y = x + 2. (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 m2= 1 m1 = 2(2) = 4 m2= 1 m1 = 4 m2= 1 tg θ= m1 - m2 . 1+ m1.m2 tg θ= 4 - 1 = 3 = 3 = 0,6 1 + (4)(1) 1 + 4 5 θ = arc tg (0,6) = 30º57'49"

Cuando: x = - 1 m1= 2x m2= 1 m1 = 2(-1) = - 2 m2= 1 m1 = - 2 m2= 1 tg θ= m1 - m2 = - 2 - (1) tg θ = - 2 - 1 = - 3 = 3 . 1+ m1.m2 1 + (-2) (1) 1 - 2 -1 θ = arc tg ( 3) θ = 71º33' 54" Puntos de intercepción Cuando x = 2 Cuando x = -1 y = x2 y = x2 y = (2)2= 4 . y = (-1)2= 1 P1 (2 , 4) P2 (-1 , 1)

Problemas - Paginas : 44 , 45 y 46

Comprobar cada una de las siguientes derivadas.

9. d (3x4 _{- 2x}2 _{+ 8)} = 12x3 - 4x dx d (3x4_{) - d (2x}2_{) + d (8)} dx dx dx 3.d (x4_{) - 2.d (x}2_{) + 0} dx dx 3 (4x3_{) - 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 _{- 5bt}3₎ = 5at4 - 15bt2. dt d (at5_{) - d (5bt}3₎ = a.d (t5) - 5b.d (t3) dt dt dt dt

a(5t4_{) - 5b (3t}2₎ = 5at4 - 15bt2 12. d ( z _{- z}2 7 ₎ = z - z6. dz 2 7 d (z 2 _{) - d ( z )} 7 = 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 2√v dx dv . dx = 1 . dv = 1 . dv 2(√v)2-1 _2(√v)1 _{dx 2√v dx} 14. 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 ( x}2₎2 _{dx x}2 _x4 = - 2 + 6 _x2 _x3 15. d (2t 4/3 _{- 3t} 2/3₎ = 8 t1/3 - 2t -1/3 dt 3 d (2t4/3_{) - d (3t}2/3₎ = 2 d (t4/3) - 3 d (t2/3) dt dt dt dt 2. 4. t4/3-1 _{- 3 . 2 . t}2/3-1 = 8 t1/3 - 2 t -1/3

3 3 . 16. 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 _{- a}2/3₎ = 2 x -1/3. dx 3 d (x2/3_{) _ d (a}2/3₎ = 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 + b x + 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 x}2 _x2 19. y = √ x - 2 ; dy = 1 + 1 . 2 √x dx 4√x x √x dy = 1 .d (√x) - (-2) . d (√x) dx 2 dx (√x)2 _dx dy = 1 .dx/dx + 2 .dx/dx = 1 . 1 + 2 . 1 = 1 + 1 dx 2 2√x x 2√x 2 2√x x 2 √x 4√x x√x

20. s = a + bt + ct2 ; ds = - a + b + 3c √ t . √t dt 2t √t 2 √t 2 s = a + bt + ct 2 = a + b. t + c.t2/2 4/2 = a + b.t1/2 + c.t3/2. √t √t √t 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 = - a + b + 3c.t 1/2 . dt 2t 2 2 2.t.t1/2 _2.t1/2 ₂ ds = _ a + b + 3c. √ t dt 2.t.√t 2.√t 2 21. y = √ax + a . ; dy = a - a .

√ax dx 2.√ax 2x.√ax

y = (ax)1/2 + a = a1/2.x1/2 + a = a1/2.x1/2 + a2/2 . (ax)1/2 _a1/2_{. x}1/2 _a1/2_{. x}1/2 y = a1/2. x1/2 + a 1/2 . x1/2 dy = d (a1/2. x1/2) + d ( a 1/2 ) = 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= a 1/2 . x -1/2 - a 1/2 . x -3/2 dx 2 2 2 2 dy = a 1/2 - a 1/2

dx 2x1/2 _2x3/2

Multiplicamos y dividimos por a1/2 _{, a cada sumando :} dy = a 1/2 .a - a1/2 1/2 .a 1/2 = a - a . dx 2x1/2_.a1/2 _2x3/2_.a1/2 _2x1/2_.a1/2 _2x.x1/2_.a1/2

dy = a - a = a - a . dx 2√x.√a 2x.√x.√a 2.√ax 2x .√ax

22. r = √1 - 2θ ; dr = - 1 . dθ √1 - 2θ r = (1 - 2θ)1/2 dr = d [(1 - 2θ)1/2] = 1 (1 - 2θ)1/2-1.d (1 - 2θ) dθ dθ 2 dθ dr = (1 - 2θ ) -1/2 .( - 2 ) = - (1 - 2θ) -1/2= - 1 = - 1 . dθ 2 (1 - 2θ)1/2 _{√1 - 2θ} 23. f ( t ) = (2 - 3t2)3 ; f '(t) = - 18t(2-3t2)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) = - 3 . (4 - 9x) 2/3 f '(x) = (4-9x)1/3 = 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 25. y = 1 ; dy = x . √a2 _{- x}2 _{dx (a}2 _{- x}2₎3/2 y = 1 = (a2 - x2)- 1/2 . (a2 _{- x}2₎1/2 dy = - 1 (a2 - x2)- 1/2 -1.d (a2 - x2) = - (a 2 - x ) 2 .(0 - 2x) .- 3/2 dx 2 dx 2 dy = -1 . (- 2.x) = x . dx 2 (a2 _{- x}2₎ 3/2 _(a2 _{- x}2₎ 3/2 26. f (θ) = (2 - 5θ)3/5 ; f '(θ) = - 3 . (2 - 5θ)2/5 f '(θ) = 3 (2 - 5θ)3/5-1 . d (2 - 5θ) 5 dθ f '(θ) = 3(2 - 5 ) θ (0 - 5) -2/5 = 3 (- 5 ) = - 3 . 5 5 (2 - 5θ)2/5 _{(2 - 5θ)}2/5

27. y = a - b 2 ; dy = 2b a - b . x dx x2 x dy = 2 ( a- b )2-1.d (a - b ) dx x dx x dy = 2 a - b . d (a) - d ( b ) = 2 a - b 0 - d (b.x-1) dx x dx dx x x dx dy = 2 a - b [- (-b.x -1-1)] dx x dy = 2 a - b [b.x-2] = 2 a - b b = 2b a - b . dx x x x2 _x2 _x 28. y = a + b 3 ; y' ₌ - 6b a + b 2 . x2 _x3 _x2 y'= 3 a + b 3-1 _{. d a + b .} x2 _{dx x}2 y'= 3a + 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' = 2a + 3bx . 2(a + bx)1/2 y = 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) (b)-1/2 + (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 30. s = t √a2 + t2 ; s'= a 2 + 2t 2 √a2 _{+ t}2 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 ds = t( a 2 + t 2 ) -1/2 .( 2.t ) + ( a2 + t2)1/2= t2 + ( a2 + t2)1/2 . dt 2 ( a2 _{+ t}2₎1/2 ds = t 2 + {(a2 + t 2 ) 1/2 } 2 = t 2 + a + t2 2= a 2 + 2t 2 .

( a2 _{+ t}2₎1/2 _{( a}2 _{+ t}2₎1/2 _{√( a}2 _{+ t}2₎

31. y = a - x ; y'= - 2a . a + x (a + x)2

(a+x).d (a-x) - (a-x).d (a+x)

dy = dx dx = (a + x) ( -1 ) - ( a - x) ( 1 ) dx (a + x)2 _{(a + x)}2 dy = - a - x - a + x = _ 2a . dx (a + x)2 _{(a + x)}2 32. y = a 2 + x 2 ; y' = 4a2 x . a2 _{- x}2 _(a2 _{- x}2₎2 (a2 _{- x}2_{).d (a}2 _{+ x}2_{) - (a}2 _{+ x}2_{).d (a}2 _{- x}2₎ dy = dx dx . dx (a2 _{- x}2₎2 dy = (a 2 - x ) (2x) - (a2 2 + x 2 ) (- 2x) = 2a 2 x - 2x + 2a3 2 x + 2x 3 dx (a2 _{- x}2₎2 _(a2 _{- x}2₎2 dy = 4a2 x . dx (a2 _{- x}2₎2 33. y = √ a 2 + x 2 ; y' = - a2 . x x2 _√a2 _{+ x}2 y = (a 2 + x 2 ) 1/2 x x.d (a2 _{+ x}2₎1/2 _{- (a}2 _{+ x}2₎1/2_{.d (x)} y'= dx dx .

x2 x. 1 . (a2 _{+ x}2₎1/2-1_{.d (a}2 _{+ x}2_{) - (a}2 _{+ x}2₎1/2₍₁₎ y'= 2 dx . x2 x(a 2 _{+ x} 2 ₎ -1/2 _{(2x) - (a}2 _{+ x}2₎1/2 _x(a 2 _{+ x} 2 ₎ -1/2 _{( 2 x) - (a}2 _{+ x}2₎1/2 y'= 2 = 2 . x2 _x2 x2 _{- (a}2 _{+ x}2₎1/2 _x 2 _{- (a + x}2 2 ₎ 1/2 _{. (a} 2 _{+ x )} 2 1/2 y'= (a 2 + x 2 ) 1/2 = (a2 + x 2 ) 1/2 . x2 _x2 x 2 _{- (a} 2 _{+ x ) x} 2 _{- a}2 2 _{- x} 2 _-a2 . y'= (a 2 + x 2 ) 1/2 = (a + x2 2 ) 1/2 = (a 2 + x ) 21/2 = x2 _x2 _x2 . 1 1 1 y'= - a2 = - a2 . x2_(a2 _{+ x}2₎1/2 _x2 _√a2 _{+ x}2 34. y = x ; y'= a2 . √a2 _{- x}2 _(a2 _{- x}2₎3/2 y = x . (a2 _{- x}2₎1/2 (a2 _{- x}2₎1/2_{.d (x) - x.d {(a}2 _{- x}2₎1/2_} y'= dx dx . {(a2 _{- x}2₎1/2_}2 (a2 _{- x}2₎1/2_{(1) - x. 1 .(a}2 _{- x}2₎1/2-1_{.d (a}2 _{- x}2_)} y'= 2 dx . (a2 _{- x}2₎2/2 (a2 _{- x}2₎1/2 _{- x.(a} 2 _{- x )} 2 _(- -1/2 _{2 x) } (a}2 _{- x}2₎1/2 _{+ x}2 _.

y'= 2 = (a2 - x ) 2 1/2 . (a2 _{- x}2₎2/2 _(a2 _{- x}2₎ (a 2 _{- x} 2 _{) . (a}1/2 2 _{- x} 2 ₎ 1/2 _{+ x} 2 _. y'= (a - x2 2 ) 1/2 = (a 2 - x 2 ) 1/2 . (a 2 - x ) 2 + x1/2 2 . (a2 _{- x}2_{) (a}2 _{- x}2_)(a2 _{- x}2₎1/2 y'= (a 2 - x 2 ) + x2 = a 2 - x 2 + x 2 = a2 . (a2 _{- x}2₎3/2 _(a2 _{- x}2₎3/2 _(a2 _{- x}2₎3/2 35. r = θ2 √3 - 4θ ; r'= 6 θ - 10θ 2 . (3 - 4θ)1/2 r = θ2 .(3 - 4θ)1/2 r'= θ2.d (3 - 4θ)1/2 + (3 - 4θ)1/2.d (θ2) dθ dθ r'= θ2. 1 .(3 - 4θ)1/2-1.d (3 - 4θ) + (3 - 4θ)1/2(2θ) 2 dθ r'= θ 2 (3 - 4 θ )-1/2 (- 4 ) + (3 - 4θ)1/2(2θ) = - 2 θ2 + (2θ)(3 - 4θ)1/2 2 (3 - 4θ)1/2 r'= - 2 θ 2 + (2θ )(3 - 4θ ) 1/2 .(3 - 4θ ) 1/2 = - 2 θ 2 + (2θ )(3 - 4 ) θ (3 - 4θ)1/2 _{(3 - 4θ)}1/2 r'= - 2 θ 2 + 6 θ - 8 θ 2 = 6 θ - 10θ 2 . (3 - 4θ)1/2 _{(3 - 4θ)}1/2 36. y = 1 - cx ; y'= - c . 1 + cx (1 + cx ) √1 - c2_x2

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_]2 (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)} dy = 2 dx 2 dx . dx (1 + cx) (1 + cx)1/2 (1 - cx) ( -c)-1/2 - (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) 2 (1 + cx )1/2 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 - c 2 _{x - c + c} 2 _x . dy = 2 (1 - cx)1/2 (1 + cx ) 1/2 = 2 (1 - cx) (1 + cx )1/2 1/2 dx ( 1 + cx ) ( 1 + cx ) . 1 1 dy = - 2 c = dx 2 (1 - cx)1/2 _{(1 + cx )}1/2_{(1 + cx )} dy = - c . dx (1 - cx)1/2 _{(1 + cx )}1/2_{(1 + cx )} dy = - c = - c . dx ( 1 + cx ) .√1 - cx . √1 + cx (1 + cx ).√(1 - cx )(1 + cx ) dy = _ c .