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If $y=(\sin x)^{(\sin x)^{(\sin x)^{-\infty}}}$, prove that $\frac{d y}{d x}=\frac{y^{2} \cot x}{(1-y \log \sin x)}$



$y=(\sin x)^{(\sin x)^{(\sin x)^{-\infty}}}$

$y=(\sin x)^{y}$

By taking log on both sides,

$\log y=\log (\sin x)^{y}$

$\log y=y(\log \sin x)$

Differentiating both sides with respect to $x$ by using product rule,

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

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

$\left(\frac{1}{y}-\log \sin x\right) \frac{d y}{d x}=\frac{y}{\sin x}(\cos x)$

$\left(\frac{1-y \log \sin x}{y}\right) \frac{d y}{d x}=y \cot x$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}^{2} \cot \mathrm{x}}{1-\mathrm{y} \log \sin \mathrm{x}}$

Hence proved.

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