Answer
\[(a)\;g'(\frac{π}{3} )=2-\sqrt 3\]
\[(b)\;h'(\frac{π}{3})=\frac{1-2\sqrt 3}{16}\]
Work Step by Step
Given that :- \[f\left(\frac{π}{3}\right)=4\;,\;f'\left(\frac{π}{3}\right)=-2\]
$ (a)\; g(x)=f(x)\;\sin x$
Differentiate with respect to $x$ using product rule
\[g'(x)=f'(x)\sin x+f(x)\cos x\]
\[g'(\frac{π}{3} )=f'(\frac{π}{3} )\sin (\frac{π}{3}) +f(\frac{π}{3} )\cos ( \frac{π}{3})\]
\[g'(\frac{π}{3} )=-2\left(\frac{\sqrt{3}}{2}\right)+(4)\left(\frac{1}{2}\right)\]
\[g'(\frac{π}{3} )=-\sqrt 3+2=2-\sqrt 3\]
Hence
\[g'(\frac{π}{3} )=2-\sqrt 3\]
$(b) \; h(x)=\large\frac{\cos x}{f(x)}$
Differentiate with respect to $x$
\[h'(x)=\frac{-\sin x\cdot \:f(x)-\cos x\: \cdot f'(x)}{(f(x))^2}\]
Using given data
\[h'(\frac{π}{3})=\frac{\frac{-\sqrt 3}{2}\times 4-\frac{1}{2}\times (-2)}{(4)^2}\]
\[h'(\frac{π}{3})=\frac{-2\sqrt 3+1}{16}\]
\[h'(\frac{π}{3})=\frac{1-2\sqrt 3}{16}\]
Hence \[h'(\frac{π}{3})=\frac{1-2\sqrt 3}{16}.\]
