## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Practice Exercises - Page 178: 55

#### Answer

a. $7$ b. $-2$ c. $\frac{5}{12}$ d. $1/4$ e. $12$ f. $9/2$ g. $3/4$

#### Work Step by Step

a. Let $h(x)=6f(x)-g(x)$, we have $h'(1)=6f'(1)-g'(1)=6(1/2)-(-4)=7$ b. Let $h(x)=f(x)g^2(x)$, we have $h'(x)=f'(x)g^2(x)+2f(x)g(x)g'(x)$ and $h'(0)=f'(0)g^2(0)+2f(0)g(0)g'(0)=-3(1^2)+2(1)(1)(1/2)=-2$ c. Let $h(x)=\frac{f(x)}{g(x)+1}$, we have $h'(x)=\frac{(g(x)+1)f'(x)-f(x)g'(x)}{(g(x)+1)^2}$ and $h'(1)=\frac{(g(1)+1)f'(1)-f(1)g'(1)}{(g(1)+1)^2}=\frac{(5+1)(1/2)-3(-4)}{(5+1)^2}=\frac{15}{36}=\frac{5}{12}$ d. Let $h(x)=f(g(x))$, we have $h'(x)=f'(g(x))g'(x)$ and $h'(0)==f'(g(0))g'(0)=f'(1)(1/2)=(1/2)(1/2)=1/4$ e. Let $h(x)=g(f(x))$, we have $h'(x)=fg(f(x))f'(x)$ and $h'(0)==g(f(0))f'(0)=g'(1)(-3)=(-4)(-3)=12$ f. Let $h(x)=(x+f(x))^{3/2}$, we have $h'(x)=\frac{3}{2}(x+f(x))^{1/2}(1+f'(x))$ and $h'(1)=\frac{3}{2}(1+f(1))^{1/2}(1+f'(1))=\frac{3}{2}(1+3)^{1/2}(1+1/2)=9/2$ g. Let $h(x)=f(x+g(x))$, we have $h'(x)=f'(x+g(x))(1+g'(x))$ and $h'(0)=f'(0+g(0))(1+g'(0))=f'(1)(1+1/2)=(1/2)(3/2)=3/4$

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