Question

Please help Suppose that f(pi/3)=4 and f'(pi/3)=-2 let g(x)=f(x)sinx and h(x)=cosx/f(x). Find g'(pi/3) and h'(pi/3)

Answer 1

Use the product rule to differentiate the first function:

$g(x)=f(x)\sin x\implies g'(x)=f'(x)\sin x+f(x)\cos x$

which means

$g'\left(\dfrac\pi3\right)=f'\left(\dfrac\pi3\right)\sin\dfrac\pi3+f\left(\dfrac\pi3\right)\cos\dfrac\pi3=2-\sqrt3$

Now use quotient rule for the second one.

$h(x)=\dfrac{\cos x}{f(x)}\implies h'(x)=\dfrac{-f(x)\sin x-f'(x)\cos x}{f(x)^2}$

so you have

$h'\left(\dfrac\pi3\right)=\dfrac{-f\left(\dfrac\pi3\right)\sin\dfrac\pi3-f'\left(\dfrac\pi3\right)\cos\dfrac\pi3}{f\left(\dfrac\pi3\right)^2}=\dfrac1{16}-\dfrac{\sqrt3}8$

Answer 2

Following are the calculation to the given question:

Given:

$\bold{f(\frac{\pi}{3})=4} \\\\\bold{f'(\frac{\pi}{3})= -2} \\\\\bold{g(x)=f(x)\ \sin x}\\\\\bold{h(x)=\frac{\cos x}{f(x)}}$

To find:

$\bold{g'(\frac{\pi}{3})=?} \\\\ \bold{h'(\frac{\pi}{3})=?}$

Solution:

$\bold{g(x) = f(x)\times \sin(x)}$

Calculating the derivative with the respect of x:

$\bold{g'(x) = f'(x)\times \sin(x) + f(x)\times \cos(x)}$

at $\ \bold{x =\frac{\pi}{3}}$

$\to \bold{g'(\frac{\pi}{3}) = f'(\frac{\pi}{3})\times (\frac{\sqrt{3}}{2}) +f(\frac{\pi}{3})\times(\frac{1}{2})}$

Calculating the value for all values of "n" $f(\frac{n}{3}) = 4$  and $f'(\frac{n}{3}) =-2$:

Therefore

$\to g'(x) = (-2)\times (\frac{\sqrt{3}}{2}) + 4\times (\frac{1}{2})\\\\\to g'(x) = -\sqrt{3} + 2\\\\Now,\\\\\to h(x) = \frac{cos(x)}{f(x)}$

Derivative with the respect of x:

$\to \bold{h'(x) = \frac{[- \sin(x)\times f(x) +f'(x)\times cos(x)]}{[f(x)]^2}}$

at $\bold{x =\frac{\pi}{3}}$  

$\to h'(\frac{pi}{3}) = [(-\frac{\sqrt{3}}{2})\times 4 + \frac{(-2)\times (\frac{1}{2})]}{[4]^2}\\\\\to h'(\frac{pi}{3}) = \frac{[-\frac{\sqrt{3}}{2} -1]}{16}\\\\\to h'(\frac{pi}{3}) = -\frac{(2+\sqrt{3})}{32}$

Learn more:

brainly.com/question/3436083