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