Question:
If $y(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}$, prove that $\frac{d y}{d x}=\frac{y^{2} \tan x}{(1-y \log \cos x)}$.
Solution:
Here,
$y=(\cos x)^{(\cos x)^{(\cos x)^{-1}}}$
$y=(\cos x)^{y}$
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,
$\frac{1}{y} \frac{d y}{d x}=y \frac{d(\log \cos x)}{d x}+\log \cos x \frac{d y}{d x}$
$\frac{1}{y} \frac{d y}{d x}=\frac{y}{\cos x} \frac{d(\cos x)}{d x}+\log \cos x \frac{d y}{d x}$
$\left(\frac{1}{y}-\log \cos x\right) \frac{d y}{d x}=\frac{y}{\cos x}(-\sin x)$
$\left(\frac{1-y \log \cos x}{y}\right) \frac{d y}{d x}=-y \tan x$
$\frac{d y}{d x}=-\frac{y^{2} \cot x}{1-y \log \cos x}$