Calculo James Stewart 3.3-3.5

Deber 3.3-3.5
Adrian Armijos (00107195) 16 de abril de 2012

3.3)Encuentre la derivada de:
9)y =
x 2 tanx x0 (2 tanx) x(2 tanx)0 (2 tanx)2 2 tanx+xsec2 x0 (2 tanx)2 sec✓ 1+sec✓ (sec✓)0 (1+sec✓)(sec✓)(1+sec✓)0 (1+sec✓)2 (sec✓tan✓)(1+sec✓) (sec✓)(sec✓tan✓) (1+sec✓)2 sec✓tan✓+sec2 ✓tan✓ sec✓ 2 tan✓ (1+sec✓)2 sec✓tan✓ (1+sec✓)2

y0 = y0 =

11)f (✓) = f 0 (✓) = f 0 (✓) = f 0 (✓) = f 0 (✓) =13)y = y0 =

senx x2 xcosx 2sinx x3

15)f (x) = exx cscx f (x)0 = e(xx csc(x) + xx csc(x)) f (x)0 = e(xx cscx + xx 19)Pruebe que
0 0

cotxcscx) = csc2 x

d dx (cotx)

29)Si H(✓) = ✓sen✓hallar H’ y H” H 0 (✓) = ✓0 sen✓ + ✓sen✓0 H 0 (✓) = sen✓ + ✓cos✓ H 00 (✓) = sen0 ✓ + (✓cos✓)0 H 00 (✓) = cos✓ + cos✓ ✓sen✓ H 00 (✓) = 2cos✓ ✓sen✓

1

30)Si f (x) = secx, hallar f 00 (⇡/4) f 0 (x) =secxtanx f 00 (x) = secxtan2 x + sec3 x f 00 (⇡/4) = sec(⇡/4)tan2 (⇡/4) + sec3 (⇡/4) p f 00 (⇡/4) = 3 2 33)¿Para que valores x de la grafica f (x) = x+2senx tiene una tangente horizontal? f 0 (x) = 1 +2cosx Para una tangente horizontal f 0 (x) = 0 1 + 2cosx = 0 cosx = 1 2 x = cos 1 1 2 x = ± 2⇡ 3 35) 39)limx!0 sen3x = x limx!0 ( 3 ) sen3x = 3limx!0 sen3x = 3 3 x 3x
tan6t 41)limt!0 sen2t =limt!0 tan6t ⇥ 6t

6t 2t

⇥

2t sen(2t)

=

1 2t 3limt!0 sen6t ⇥ limt!0 cos6t ⇥ limt!0 sen(2t) = 3 ⇥ 13 = 3 6t

3.4 Regla de la cadena
4)y = tan(senx) y 0 = sec2 (senx)(cosx) 11)g(t) = g(t) =g(t) = g(t) =
1 (t4 +1)3

3(t4 +1)0 (t4 +1)4 3(4t3 ) (t4 +1)4 12t3 (t4 +1)4 kx kx

15)y = xe

y = x(e kx ( k)) + e = e kx ( kx + 1)

(1)

2

19)y = (2x y0 = y0 =

5)4 (8x2

5)

34(2x 5)3 (2) (2x 5)4 (16x) (8x2 5)3 + 3(8x2 5)4 8(2x 5)3 48x(2x 5)4 (8x2 5)3 + (8x2 5)4

23)y = excosx y 0 = excosx (xcosx)0 y 0 = excosx (cosx
y 30)G(y) = ( y+1 )5 y y G(y)0 = 5( y+1 )4 ( y+1)0 y G(y)0 = 5( y+1 )4 (2y(y + 1) y G(y)0 = 5( y+1 )4 (2y(y + 1)
2 2 2 2 2

xsenx)

1 1

+ y 2 ( 1(y + 1) y 2 (y + 1)
2

2

)(1)
2

y ) = 5( y+1 )4 ( 2y21 +2y

1 ) (y 3 +y 2 )2...