668 Appendix C 11 Answers & Hints for Selected Exercises- d~fo(goh) = J'[(goh)(x)]·(goh)'(x) = J'[(goh)(x)]· g'(h(x))·h'(x) =
(!' o go h)(x) · (g' o h)(x) · h'(x). - For x > 0,
1 1
d~ [ x + ( x + y'x)! ]
2
= ~ [ x + ( x + y'x)! ] -2
[ 1 + ~ ( x + y'x)-! d~ ( x + y'x)]
1+-1-
1 + 2-/X
2Jx + v'x 2Jx+v'x+l+~4Jx + vxJx + Jx + v'x
1+2y'x + 4Jx^2 + xy'x
8y'xJx + vxJx + Jx + v'x
- Let a> 0, a=/= 1, and f(x) =ax. By Exercise 5.6.16, f(x) =ex Ina.
(a) By Thm. 6.2.9 and the chain rule, f'(x) =ex Ina Ina= ax Ina.
(b) By Thm. 5.6. 25 , loga x = :~~, so by Thm. 6.2.9 and the chain rule,
A.1 dx oga X -.!.- x ._l In __ a - x In l a. - Use Ex. 12 and the chain rule.
- Suppose f , g are differentiable everywhere and f > 0. Let u = f(x)g(x).
Then ln u = g( x) ln f ( x) is differentiable, and taking the derivative of both
sides,
~ ~~ = g(x) · ~g] + g'(x) lnf(x), so ~~ = u [g(x) · ~g} + g'(x) lnf(x)].
1 7. (a) For x =/= 0, d~ x^2 sin ~ = x^2 (cos ~) ( -:Iz-) + 2x sin ~ = 2x sin ~ - cos ~.
(b) f' ( x) is continuous whenever x =/= 0, but lim f' ( x) = lim (2x sin -x^1 - cos .!. ) ,
x-o x-o x
which does not exist for the following reason :
Let Xn = - (^2) ~^1. Then Xn , 0 and n-oo lim f'(xn) = n-oo lim (0 - 1) = -1.
Let Yn = ""+ 2 12 7rn. Then Yn , 0 but n-+oo lim f'(Yn) = n-+oo lim (2yn -0) = 0.
Since n-+oo lim f^1 ( Xn) =/= n-+oo lim f^1 (Yn), x -+O lim f' ( x) does not exist.
- Suppose f is p eriodic with period p > 0 and differentiable on [a , a+ p).
(a) Let xo ER Let k be the smallest integer 3 x 0 +kp 2:: a. Then x 0 +kp E
[a, a+ p) and, letting u = x + kp, uo = xo + kp,
f'(xo) = lim f(x) - f(xo) = lim f(x + kp) - f(x^0 + kp)
x-xo x -x 0 x-xa (x + kp) - (x 0 - kp)
= lim f(u) - f(uo) = f'(u), which exists since uo E [a , a+ p).
u-uo U - Uo
(b) f'(x + p) lim f(x + p + h) - f(x + p) lim f(x + h) - f(x)
h-o h h-o h
f'(x).
EXERCISE SET 6.3- Revise the proof of Thm. 6.3.2, making appropriate changes.