Question

Mark the correct alternative in each of the following:

If f(x) = x sinx, then $f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=$

(a) 0 (b) 1 (c) −1 (d) $\frac{1}{2}$

Answer 1

f(x) = x sinx

Differentiating both sides with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=x\times \frac{d}{dx}\left(\sin x\right)+\sin x\times \frac{d}{dx}\left(x\right)\left(\mathrm{Product}\mathrm{rule}\right) \\ =x\times \cos x+\sin x\times 1 \\ =x\cos x+\sin x \end{gathered}$$

Putting $x=\frac{\mathrm{\pi }}{2}$, we get

$$\begin{gathered} f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{2}\times \cos \left(\frac{\mathrm{\pi }}{2}\right)+\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ =\frac{\mathrm{\pi }}{2}\times 0+1 \\ =1 \end{gathered}$$

Hence, the correct answer is option (b).