Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 32

If $\displaystyle f \left( \frac{\pi}{3} \right) = 4$ and $\displaystyle f'\left( \frac{\pi}{3} \right) = -2$, and let $g(x) = f(x) \sin x$ and $\displaystyle h(x) = \frac{\cos x}{f(x)}$.

Find (a) $\displaystyle g'\left( \frac{\pi}{3} \right)$ and (b) $\displaystyle h' \left( \frac{\pi}{3} \right)$.

a. ) $g(x) = f(x) \sin x$


$\begin{equation} \begin{aligned} g'(x) =& f(x) \frac{d}{dx} (\sin x) + (\sin x) \frac{d}{dx} (f(x)) && \text{Using Product Rule} \\ \\ g'(x) =& f(x) \cos x + f'(x) \sin x && \text{Substitute the given value} \\ \\ g'\left( \frac{\pi}{3} \right) =& f\left( \frac{\pi}{3} \right) \cos \left( \frac{\pi}{3} \right) + f' \left( \frac{\pi}{3} \right) \sin \left( \frac{\pi}{3} \right) && \text{Simplify the equation} \\ \\ g'\left( \frac{\pi}{3} \right) =& -(4)\left( \frac{1}{2} \right) + (-2) \left( \frac{\sqrt{3}}{2} \right) && \text{Simplify the equation} \\ \\ g' \left( \frac{\pi}{3} \right) =& 2 - \sqrt{3} && \end{aligned} \end{equation}$



b.) $\displaystyle h(x) = \frac{\cos x}{f(x)}$



$\begin{equation} \begin{aligned} h'(x) =& \frac{\displaystyle f(x) \frac{d}{dx} (\cos x) - \left[ \cos x \frac{d}{dx} f(x) \right]}{(f(x))^2} && \text{Apply Quotient Rule} \\ \\ h'(x) =& \frac{f(x) (- \sin x) - f'(x) \cos x}{(f(x))^2} && \text{Substitute given value} \\ \\ h'(x) =& \frac{\displaystyle -f \left( \frac{\pi}{3} \right) \sin \left( \frac{\pi}{3} \right) - f'\left( \frac{\pi}{3} \right) \cos \left( \frac{\pi}{3} \right)}{ \displaystyle \left( f\left( \frac{\pi}{3} \right)\right)^2} && \text{Simplify the equation} \\ \\ h'\left( \frac{\pi}{3} \right) =& \frac{\displaystyle (4) \left( \frac{\sqrt{3}}{2} \right) - (-2) \left( \frac{1}{2} \right)}{(4)^2} && \text{Simplify the equation} \\ \\ h'\left( \frac{\pi}{3} \right) =& \frac{-2 \sqrt{3} + 1}{16} && \text{Simplify the equation} \\ \\ h'\left( \frac{\pi}{3} \right) =& \frac{1}{16} - \frac{\sqrt{3}}{8} && \text{} \\ \\ \end{aligned} \end{equation}$