Answer
a. $\frac{2}{3}$
b. $2\pi+5$
c. $15-8\pi$
d. $\frac{37}{6}$
e. $-1$
f. $\frac{\sqrt 2}{24}$
g. $\frac{5}{32}$
h. $\frac{-5}{3\sqrt 17}$
Work Step by Step
a. let $y$ = $2f(x)$
$y'$ = $2f'(x)$
$2f'(2)$ = $(2)(\frac{1}{3})$ = $\frac{2}{3}$
b. let $y$ = $f(x)+g(x)$
$y'$ = $f'(x)+g'(x)$
$f'(3)+g'(3)$ = $2\pi+5$
c. let $y$ = $f(x)$$g(x)$
$y'$ = $f(x)g'(x)+g(x)f'(x)$
$f(3)g'(3)+g(3)f'(3)$ = $(3)(5)+(-4)(2\pi)$ = $15-8\pi$
d. let y = $\frac{f(x)}{g(x)}$
$y'$ = $\frac{g(x)f'(x)-f(x)g'(x)}{g^{2}(x)}$
$\frac{g(2)f'(2)-f(2)g'(2)}{g^{2}(2)}$ = $\frac{(2)(\frac{1}{3})-(8)(-3)}{(2)^{2}}$ = $\frac{74}{12}$ = $\frac{37}{6}$
e. let $y$ = $f(g(x))$
$y'$ = $f'(g(x))g'(x)$
$f'(g(x))g'(x)$ = $f'(g(2))g'(2)$ = $f'(2)[-3]$ = $(\frac{1}{3})(-3)$ = $-1$
f. let $y$ = $\sqrt {f(x)}$
$y'$ = $\frac{1}{2\sqrt {f(x)}}f'(x)$
$\frac{1}{2\sqrt {f(x)}}f'(x)$ = $\frac{1}{2\sqrt {8}}(\frac{1}{3})$ = $\frac{1}{12\sqrt 2}$ = $\frac{\sqrt 2}{24}$
g. let $y$ = $\frac{1}{g^{2}(x)}$
$y'$ = $\frac{-2g(x)g'(x)}{g^{3}(x)}$
$\frac{-2g'(x)}{g^{3}(x)}$ = $\frac{-2g'(3)}{g^{3}(3)}$ = $\frac{-2(5)}{(-4)^{3}}$ = $\frac{5}{32}$
h. let $y$ = $\sqrt {f^{2}(x)+g^{2}(x)}$
$y'$ = $\frac{2f(x)f'(x)+2g(x)g'(x)}{2\sqrt {f^{2}(x)+g^{2}(x)}}$ = $\frac{f(x)f'(x)+g(x)g'(x)}{\sqrt {f^{2}(x)+g^{2}(x)}}$
$\frac{f(x)f'(x)+g(x)g'(x)}{\sqrt {f^{2}(x)+g^{2}(x)}}$ = $\frac{f(2)f'(2)+g(2)g'(2)}{\sqrt {f^{2}(2)+g^{2}(2)}}$ = $\frac{(8)(\frac{1}{3})+(2)(-3)}{\sqrt {(8)^{2}+(2)^{2}}}$ = $\frac{-5}{3\sqrt 17}$
