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
