Question

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

Answer 1

$\mathrm{We}\mathrm{have},y={\left(\cos x\right)}^{{\left(\cos x\right)}^{{\left(\cos x\right)}^{....\infty }}}$
$\Rightarrow y={\left(\cos x\right)}^y$
Taking log on both sides,
$$\begin{gathered} \log y=\log {\left(\cos x\right)}^y \\ \Rightarrow \log y=y\log \left(\cos x\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\left\{\log \cos x\right\}+\log \cos x\frac{dy}{dx} \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=y\left(\frac{1}{\cos x}\right)\frac{d}{dx}\left(\cos x\right)+\log \cos x\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(\frac{1}{y}-\log \cos x\right)=\frac{y}{\cos x}\left(-\sin x\right) \\ \Rightarrow \frac{dy}{dx}\left(\frac{1-y\log \cos x}{y}\right)=-y\tan x \\ \Rightarrow \frac{dy}{dx}=-\frac{y^2\tan x}{\left(1-y\log \cos x\right)} \end{gathered}$$