5. if ( f(x)=sin left(x^{2}+pi\right) ), then ( f^{prime}(sqrt{2 pi})= )\n(a) ( -2 sqrt{2 pi} ) (b) ( -2 )…

Explanation:

Step1: Find the derivative of (f(x))

Use the chain rule. If (y = \sin(u)) and (u=x^{2}+\pi), then (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). We know that (\frac{d}{du}(\sin(u))=\cos(u)) and (\frac{d}{dx}(x^{2}+\pi)=2x). So (f^{\prime}(x)=\cos(x^{2}+\pi)\cdot2x).

Step2: Substitute (x = \sqrt{2\pi}) into (f^{\prime}(x))

First, calculate (x^{2}) when (x=\sqrt{2\pi}), (x^{2}=2\pi). Then (f^{\prime}(\sqrt{2\pi})=\cos(2\pi+\pi)\cdot2\sqrt{2\pi}). Since (\cos(2\pi +\alpha)=\cos(\alpha)) and (\cos(3\pi)=\cos(\pi)= - 1). So (f^{\prime}(\sqrt{2\pi})=-1\times2\sqrt{2\pi}=-2\sqrt{2\pi}).

Answer:

A. (-2\sqrt{2\pi})