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If $y(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}$, prove that $\frac{d y}{d x}=\frac{y^{2} \tan x}{(1-y \log \cos x)}$.



$y=(\cos x)^{(\cos x)^{(\cos x)^{-1}}}$

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

By taking log on both sides,

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

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

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

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

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

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

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

$\frac{d y}{d x}=-\frac{y^{2} \cot x}{1-y \log \cos x}$

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