Question

If $y=(cosx{)}^{(cosx{)}^{{cosx}^{-\infty }}}$, then show that $\frac{dy}{dx}=\frac{y^{2}tanx}{ylog\left(cosx\right)-1}$

Answer 1

We have, $y=(cosx{)}^{(cosx{)}^{{cosx}^{-\infty }}}$

$\Rightarrow y=(cosx{)}^y\therefore logy=log(cosx{)}^y\Rightarrow logy=ylog\left(cosx\right)$

On differentiating w.r.t. x, we get

$\frac{1}{y}.\frac{dy}{dx}=y.\frac{d}{dx}\left(log\right(cosx\left)\right)+log\left(cosx\right).\frac{dy}{dx}\Rightarrow \frac{1}{y}.\frac{dy}{dx}=\frac{y}{cosx}.\frac{d}{dx}cosx+log\left(cosx\right).\frac{dy}{dx}\Rightarrow \frac{dy}{dx}[\frac{1}{y}-log(cosx\left)\right]=\frac{-ysinx}{cosx}=-ytanx\therefore \frac{dy}{dx}=\frac{-y^{2}tanx}{(1-ylog(cosx\left)\right)}=\frac{y^{2}tanx}{ylog\left(cosx\right)-1}$