question 5\nsuppose that $f(\\frac{\\pi}{6})=-2$ and $f(\\frac{\\pi}{6}) = \\sqrt{3}$, and let…

Explanation:

Step1: Find $g'(x)$ 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 $v=\cos x$. So, $g'(x)=f'(x)\cos x - f(x)\sin x$.

Step2: Evaluate $g'\left(\frac{\pi}{6}\right)$

Substitute $x = \frac{\pi}{6}$ into $g'(x)$. We know that $f\left(\frac{\pi}{6}\right)=-2$, $f'\left(\frac{\pi}{6}\right)=\sqrt{3}$, $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$, and $\sin\frac{\pi}{6}=\frac{1}{2}$. [ \begin{align*} g'\left(\frac{\pi}{6}\right)&=f'\left(\frac{\pi}{6}\right)\cos\frac{\pi}{6}-f\left(\frac{\pi}{6}\right)\sin\frac{\pi}{6}\ &=\sqrt{3}\times\frac{\sqrt{3}}{2}-(-2)\times\frac{1}{2}\ &=\frac{3}{2}+ 1\ &=\frac{3 + 2}{2}=\frac{5}{2} \end{align*} ]

Step3: Find $h'(x)$ 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, $u = f(x)$ and $v=\tan x$. So, $h'(x)=\frac{f'(x)\tan x - f(x)\sec^{2}x}{\tan^{2}x}$.

Step4: Evaluate $h'\left(\frac{\pi}{6}\right)$

We know that $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$, $\sec\frac{\pi}{6}=\frac{2}{\sqrt{3}}$, $f\left(\frac{\pi}{6}\right)=-2$, and $f'\left(\frac{\pi}{6}\right)=\sqrt{3}$. [ \begin{align*} h'\left(\frac{\pi}{6}\right)&=\frac{f'\left(\frac{\pi}{6}\right)\tan\frac{\pi}{6}-f\left(\frac{\pi}{6}\right)\sec^{2}\frac{\pi}{6}}{\tan^{2}\frac{\pi}{6}}\ &=\frac{\sqrt{3}\times\frac{\sqrt{3}}{3}-(-2)\times\frac{4}{3}}{\frac{1}{3}}\ &=\frac{1+\frac{8}{3}}{\frac{1}{3}}\ &=\frac{\frac{3 + 8}{3}}{\frac{1}{3}}\ &=11 \end{align*} ]

Answer:

$g'\left(\frac{\pi}{6}\right)=\frac{5}{2}$, $h'\left(\frac{\pi}{6}\right)=11$