If $y=x^{(x \sin x)}$ then $\frac{d y}{d x}=?$

$x^{(x \sin x)}[\sin (1+\log x)+x \log x \cos x]$

$y=x^{x \sin x}$         ....(1)

taking logarithm on both sides

$\log y=\log x^{x \sin x} \Rightarrow \log y=x \sin x \log x$     (as $\log a^b=b \log a$)

differentiating both sides w.r.t (x)

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(x \sin x \log x)$

$\frac{1}{4} \frac{d y}{d x}=x \sin x \frac{d \log x}{d x}+x \log x \frac{d \sin x}{d x}+\sin x \log x \frac{d x}{d x}$

[using product rule for 3 variables

(UVW)' = U'(VW) + U(VW)'

= U'VW + UV'W + UVW']

so $\frac{d y}{d x}=y\left[\frac{x}{x} \sin x+x \log x \cos x+\sin x \log x\right]$

from (1)

$=y(\sin x(1+\log x)+x \log x \cos x)$

$\frac{d y}{d x}=x^{x \sin x} (\sin x(1+\log x)+x \log x \cos x)$