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)

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

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

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