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Question:

If $y=\sin \left(x^{x}\right)$, prove that $\frac{d y}{d x}=\cos \left(x^{x}\right) \cdot x^{x}(1+\log x)$

Solution:

Here,

$y=\sin \left(x^{x}\right) \ldots \ldots$ (i)

Let $u=x^{x} \ldots \ldots$ (ii)

Taking log on both sides,

$\log u=\log x^{x}$

$\log u=x \log x$

Differentiating both sides with respect to $x$,

$\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x} \log \mathrm{x})$

$=x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(x)$

$=x\left(\frac{1}{x}\right)+\log x(1)$

$\frac{1}{u} \frac{d u}{d x}=1+\log x$

$\frac{d u}{d x}=u(1+\log x)$

$\frac{d u}{d x}=x^{x}(1+\log x) \ldots \ldots$ (iii) [from (ii)]

Now, using equation (ii) in (i)

$y=\sin u$

Differentiating both sides with respect to $\mathrm{x}$,

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

$=\cos u \frac{d u}{d x}$

Using equation (ii) and (iii),

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

Hence Proved.