suppose $f(\\frac{\\pi}{3}) = 2$ and $f(\\frac{\\pi}{3})=-7$. let $g(x)=f(x)\\sin(x)$ and $h(x)=\\frac{\\cos(…

Explanation:

Step1: Find the derivative of (g(x)) using product - rule

The product - rule states that if (g(x)=u(x)v(x)), then (g'(x)=u'(x)v(x)+u(x)v'(x)). Here, (u = f(x)) and (v=\sin(x)), so (g'(x)=f'(x)\sin(x)+f(x)\cos(x)).

Step2: Evaluate (g'\left(\frac{\pi}{3}\right))

Substitute (x = \frac{\pi}{3}) into (g'(x)). We know that (f\left(\frac{\pi}{3}\right)=2), (f'\left(\frac{\pi}{3}\right)=-7), (\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}), and (\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}). [ \begin{align*} g'\left(\frac{\pi}{3}\right)&=f'\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{3}\right)+f\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{3}\right)\ &=-7\times\frac{\sqrt{3}}{2}+2\times\frac{1}{2}\ &=1-\frac{7\sqrt{3}}{2} \end{align*} ]

Step3: Find the derivative of (h(x)) using quotient - rule

The quotient - rule states that if (h(x)=\frac{u(x)}{v(x)}), then (h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v^{2}(x)}). Here, (u = \cos(x)) and (v = f(x)), so (h'(x)=\frac{-\sin(x)f(x)-\cos(x)f'(x)}{f^{2}(x)}).

Step4: Evaluate (h'\left(\frac{\pi}{3}\right))

Substitute (x=\frac{\pi}{3}) into (h'(x)). We know that (f\left(\frac{\pi}{3}\right)=2), (f'\left(\frac{\pi}{3}\right)=-7), (\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}), and (\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}). [ \begin{align*} h'\left(\frac{\pi}{3}\right)&=\frac{-\sin\left(\frac{\pi}{3}\right)f\left(\frac{\pi}{3}\right)-\cos\left(\frac{\pi}{3}\right)f'\left(\frac{\pi}{3}\right)}{f^{2}\left(\frac{\pi}{3}\right)}\ &=\frac{-\frac{\sqrt{3}}{2}\times2-\frac{1}{2}\times(-7)}{2^{2}}\ &=\frac{- \sqrt{3}+\frac{7}{2}}{4}\ &=\frac{7 - 2\sqrt{3}}{8} \end{align*} ]

Answer:

(a) (1-\frac{7\sqrt{3}}{2}) (b) (\frac{7 - 2\sqrt{3}}{8})