Question

If $y=x^{\sin x}+(\sin x)^{x}$ then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is

• A. $\frac{4}{\pi}$
• B. $1$
• C. $\pi \log \frac{\pi}{2}$
• D. $\frac{\pi^{2}}{2}$

Answer 1

Answer: (B)

$y=x^{\sin x}+(\sin x)^{x}$
$\frac{d y}{d x}=\left[x^{\sin x}\right]\left[\frac{\sin x}{x}+\cos x \cdot \log x\right]+$
$(\sin x)^{x}[x \cos x+\log \sin x]$
$x=\frac{\pi}{2}$
$=\frac{\pi}{2}\left[\frac{2}{\pi}\right]+1[0+0]=1$