# Differentiate $$x^{sinx}$$ with respect to x.

## Solution :

Let y = $$x^{sinx}$$. Then,

Taking log both sides,

log y = sin x.log x

$$\implies$$ y = $$e^{sin x.log x}$$

By using logarithmic differentiation,

On differentiating both sides with respect to x, we get

$$dy\over dx$$ = $$e^{sin x.log x}$$$$d\over dx$$(sin x.log x)

$$\implies$$ $$dy\over dx$$ = $$x^{sin x}{log x {d\over dx}(sin x) + sin x {d\over dx}(log x)}$$

$$\implies$$ $$dy\over dx$$ = $$x^{sin x}(cos x.log x + {sin x\over x}$$)