Answer
a. $1$
b. $6$
c. $1$
d. $-\frac{1}{9}$
e. $-\frac{40}{3}$
f. $-\frac{1}{3}$
g. $-\frac{4}{9}$
Work Step by Step
a. Let $h(x)=5f(x)-g(x)$, we have at $x=1$, $h'(x)=5f'(x)-g'(x)=5f'(1)-g'(1)=5(-1/3)-(-8/3)=1$
b. At $x=0$, $(f(x)g^3(x))'=f'(x)g^3(x)+3f(x)g^2(x)g'(x)=f'(0)g^3(0)+3f(0)g^2(0)g'(0)=5(1)^3+3(1)(1)^2(1/3)=5+1=6$
c. Let $k(x)=\frac{f(x)}{g(x)+1}$, at $x=1$, we have $k'(x)=\frac{(g(x)+1)f'(x)-f(x)g'(x)}{(g(x)+1)^2}=\frac{(g(1)+1)f'(1)-f(1)g'(1)}{(g(1)+1)^2}=\frac{(-4+1)(-1/3)-(3)(-8/3)}{(-4+1)^2}=\frac{1+8}{9}=1$
d. At $x=0$, $(f(g(x)))'=f'(g(0))g'(0)=f'(1)(1/3)=(-1/3)(1/3)=-\frac{1}{9}$
e. At $x=0$, $(g(f(x)))'=g'(f(0))f'(0)=g'(1)(5)=(-8/3)(5)=-\frac{40}{3}$
f. Let $L(x)=(x^{11}+f(x))^{-2}$, at $x=1$, $L'(x)=-2(x^{11}+f(x))^{-3}(11x^{10}+f'(x))=-2(1^{11}+f(1))^{-3}(11(1)^{10}+f'(1))=-2(1+3)^{-3}(11-1/3)=-2(4)^{-3}(32/3)=-\frac{1}{3}$
g. At $x=0$, $(f(x+g(x)))'=f'(x+g(x))(1+g'(x))=f'(0+g(0))(1+g'(0))=f'(1)(1+1/3)=(-1/3)(4/3)=-\frac{4}{9}$
