Answer
(a) $-6\;\;,$ (b) 24 $\;\;$ (c) 4$\;\;,$ (d) $\frac{-36}{49}$
Work Step by Step
Given that $f(4)=2\,,\,g(4)=5\:,\,f'(4)=6\,,\,g'(4)=-3$
(a) $h(x)=3f(x)+8g(x)$
Differentiating with respect to $x$
$\Rightarrow h'(x)=3f'(x)+8g'(x)$
$\Rightarrow h'(4)=3f'(4)+8g'(4)$
Using given that
$\Rightarrow h'(4)=3(6)+8(-3)$
$\Rightarrow h'(4)=18-24=-6$
(b) $h(x)=f(x)g(x)$
Differentiating with respect to $x$ using product rule
$\Rightarrow h'(x)=f'(x)g(x)+f(x)g'(x)$
$\Rightarrow h'(4)=f'(4)g(4)+f(4)g'(4)$
Using given that
$\Rightarrow h'(4)=(6)(5)+(2)(-3)$
$\Rightarrow h'(4)=30-6=24$
(c) $\;h(x)=\frac{f(x)}{g(x)}$
Differentiating with respect to $x$ by quotient rule
$\Rightarrow h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$
$\Rightarrow h'(4)=\frac{f'(4)g(4)-f(4)g'(4)}{[g(4)]^2}$
$\Rightarrow h'(4)=\frac{(6)(5)-(2)(-3)}{(-3)^2}$
$\Rightarrow h'(4)=\frac{30+6}{9}=\frac{36}{9}=4$
(d) $h(x)=\frac{g(x)}{f(x)+g(x)}$
Differentiating with respect to $x$ using quotient rule
$\Rightarrow h'(x)=\frac{g'(x)[f(x)+g(x)]-g(x)[f(x)+g(x)]'}{[f(x)+g(x)]^2}$
$\Rightarrow h'(x)=\frac{g'(x)[f(x)+g(x)]-g(x)[f'(x)+g'(x)]}{[f(x)+g(x)]^2}$
$\Rightarrow h'(x)=\frac{g'(x)f(x)+g'(x)g(x)-g(x)f'(x)-g(x)g'(x)}{[f(x)+g(x)]^2}$
$\Rightarrow h'(x)=\frac{g'(x)f(x)-g(x)f'(x)}{[f(x)+g(x)]^2}$
$\Rightarrow h'(4)=\frac{g'(4)f(4)-g(4)f'(4)}{[f(4)+g(4)]^2}$
Using given data
$\Rightarrow h'(4)=\frac{(-3)(2)-(5)(6)}{(2+5)^2}$
$\Rightarrow h'(4)=\frac{-6-30}{49}=\frac{-36}{49}$ .
