Answer
a) $2/3$
b) $2\pi+5$
c) $-8\pi+15$
d) $37/6$
e) $-1$
f) $\sqrt2/24$
g) $5/32$
h) $-5/3\sqrt17$
Work Step by Step
a) The derivative of $2f(x)$: $$(2f(x))'=2f'(x)$$
For $x=2$: $$2f'(2)=2\times\frac{1}{3}=\frac{2}{3}$$
b) The derivative of $f(x)+g(x)$: $$(f(x)+g(x))'=f'(x)+g'(x)$$
For $x=3$: $$f'(3)+g'(3)=2\pi+5$$
c) The derivative of $f(x)g(x)$: $$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$$
For $x=3$: $$f'(3)g(3)+f(3)g'(3)=2\pi(-4)+3\times5=-8\pi+15$$
d) $$\Big(\frac{f(x)}{g(x)}\Big)'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$$
Therefore, $$\Big(\frac{f(2)}{g(2)}\Big)'=\frac{f'(2)g(2)-f(2)g'(2)}{g^2(2)}=\frac{\frac{1}{3}\times2-8(-3)}{2^2}=\frac{\frac{2}{3}+24}{4}=\frac{37}{6}$$
e) $$\Big(f(g(x))\Big)'=f'(g(x))\times g'(x)$$
Therefore, $$\Big(f(g(2))\Big)'=f'(g(2))\times g'(2)=f'(2)\times(-3)=\frac{1}{3}\times(-3)=-1$$
f) $$\Big(\sqrt{f(x)}\Big)'=\Bigg[\Big(f(x)\Big)^{1/2}\Bigg]'=\frac{1}{2}\Big(f(x)\Big)^{-1/2}f'(x)=\frac{f'(x)}{2\sqrt{f(x)}}$$
Therefore, $$\Big(\sqrt{f(2)}\Big)'=\frac{f'(2)}{2\sqrt{f(2)}}=\frac{\frac{1}{3}}{2\sqrt8}=\frac{1}{6\sqrt8}=\frac{1}{12\sqrt2}=\frac{\sqrt2}{24}$$
g) $$\Big(\frac{1}{g^2(x)}\Big)'=\frac{-1\Big(g^2(x)\Big)'}{g^4(x)}=-\frac{2g(x)g'(x)}{g^4(x)}$$
Therefore, $$\Big(\frac{1}{g^2(3)}\Big)'=-\frac{2g(3)g'(3)}{g^4(3)}=-\frac{2\times(-4)\times5}{(-4)^4}=-\frac{(-40)}{256}=\frac{5}{32}$$
h) $$\Big(\sqrt{f^2(x)+g^2(x)}\Big)'=\frac{1}{2\sqrt{f^2(x)+g^2(x)}}(f^2(x)+g^2(x))'=\frac{1}{2\sqrt{f^2(x)+g^2(x)}}(2f(x)f'(x)+2g(x)g'(x))$$ $$=\frac{f(x)f'(x)+g(x)g'(x)}{\sqrt{f^2(x)+g^2(x)}}$$
Therefore, $$\Big(\sqrt{f^2(2)+g^2(2)}\Big)'=\frac{f(2)f'(2)+g(2)g'(2)}{\sqrt{f^2(2)+g^2(2)}}=\frac{8\times\frac{1}{3}+2\times(-3)}{\sqrt{8^2+2^2}}$$ $$=\frac{\frac{8}{3}-6}{\sqrt{68}}=\frac{-\frac{10}{3}}{2\sqrt{17}}=-\frac{5}{3\sqrt{17}}$$
