Differentiate x^ x , x > 0 w.r.t. x. - Vidyadip

Question

Differentiate $x^{\sin x}, x > 0 w.r.t. x$.

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Answer

Let $y = x^{\sin x}$.
Taking logarithm on both sides, we have
log y = $sinx\times logx$
Therefore $\frac{1}{y} \cdot \frac{d y}{d x}=\sin x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(\sin x)$
or $\frac{1}{y} \frac{d y}{d x}=(\sin x) \frac{1}{x}$ + log x cos x
or $\frac{d y}{d x}=y\left[\frac{\sin x}{x}+\cos x \log x\right]$
or $\frac{d y}{d x}$ = $x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]$
or $\frac{d y}{d x}$ =$ x ^{sinx-1}$ sin $x + x^{\sin x}. cos\ x\ log\ x$

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