If $y = {(\cos x)^{{{(\cos x)}^{{{(\cos x)}^{ - \infty }}}}}},$ then show that $\frac{{dy}}{{dx}} = \frac{{{y^2}\tan x}}{{y\log \cos x - 1}}.$

We have, $y = {(\cos x)^{{{(\cos x)}^{(\cos x) \cdots \infty }}}}$

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

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

$\Rightarrow$ $\log y = y\log \cos x$

On differentiating w.r.t. $x$, we get
$\frac{1}{y} \cdot \frac{{dy}}{{dx}} = y \cdot \frac{d}{{dx}}\log \cos x + \log \cos x \cdot \frac{{dy}}{{dx}}$

$\Rightarrow$ $\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \frac{y}{{\cos x}} \cdot \frac{d}{{dx}}\cos x + \log \cos x \cdot \frac{{dy}}{{dx}}$
$\Rightarrow$ $\frac{{dy}}{{dx}}\left[ {\frac{1}{y} - \log \cos x} \right] = \frac{{ - y\sin x}}{{\cos x}} = - y\tan x$

therefore,$\frac{{dy}}{{dx}} = \frac{{ - {y^2}\tan x}}{{(1 - y\log \cos x)}}$