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

Given, \(y=x^{\sin x}+(\sin x)^x\)
Let \(u=x^{\sin x}\) and \(v=(\sin x)^x\)
Now, \(u=x^{\sin x}, \log u=\sin x \log x\)
Differentiating w.r.t. \(x\), we get
\[
\begin{aligned}
& \frac{1}{u} \cdot \frac{d u}{d x}=\sin x \cdot \frac{1}{x}+\log x \times \cos x \\
& \frac{d u}{d x}=u\left(\frac{\sin x}{x}+\cos x \log x\right) \\
& \frac{d u}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x \log x\right)
\end{aligned}
\]
Now, \(v=(\sin x)^x \Rightarrow \log v=x \log (\sin x)\)
Differentiating w.r.t. \(x\), we get
\[
\frac{1}{v} \frac{d v}{d x}=x \frac{\cos x}{\sin x}+\log (\sin x)
\]
\(\begin{aligned} & \frac{d v}{d x}=v(x \cot x+\log \sin x) \\ & \frac{d v}{d x}=(\sin x)^x(x \cot x+\log \sin x)\end{aligned}\)

Adding Eqs. (i) and (ii), we get
\[
\begin{aligned}
& \frac{d u}{d x}+\frac{d v}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\right.+\cos x \log x) \\
&+(\sin x)^x(x \cot x+\log \sin x) \\
& \frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x \log x\right) \\
&++(\sin x)^x(x \cot x+\log \sin x)
\end{aligned}
\]
At \(x=\frac{\pi}{2}\)
\[
\begin{aligned}
\frac{d y}{d x} & =\left(\frac{\pi}{2}\right)\left(\frac{1}{\pi / 2}+0 \log \frac{\pi}{2}\right)+(1)^{\frac{\pi}{2}}\left(\frac{\pi}{2} \cdot 0+\log (1)\right) \\
\frac{d y}{d x} & =\left(\frac{\pi}{2}\right)\left(\frac{2}{\pi}+0\right)+1 \cdot(0+0)=1
\end{aligned}
\]