Question

If $f\left(x\right)=xsinx,$ then $f^{\prime }\frac{\pi }{2}$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$1$

Given that:

$f\left(x\right)=x\sin x........\left(1\right)$

Differentiating equation with respect to $x$.

$\frac{df\left(x\right)}{dx}=\frac{d\left(x\sin x\right)}{dx}$

$f^{\prime }\left(x\right)=\frac{d\left(x\right)}{dx}\sin x+x\frac{d\left(\sin x\right)}{dx}$

$f^{\prime }\left(x\right)=\sin x+x\cos x$

Now puting $x=\frac{\pi }{2}$

$f^{\prime }\left(\frac{\pi }{2}\right)=\sin \frac{\pi }{2}+\frac{\pi }{2}\cos \frac{\pi }{2}$

$\because \sin \frac{\pi }{2}=1,\cos \frac{\pi }{2}=0$

$f^{\prime }\left(\frac{\pi }{2}\right)=1$