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

\[\text{ We have, y} = \left( \cos x \right)^{\left( \cos x \right)^{\left( \cos x \right)^{. . . . \infty}}} \]

\[\Rightarrow y = \left( \cos x \right)^y\]

Taking log on both sides,

\[\log y = \log \left( \cos x \right)^y \]
\[ \Rightarrow \log y = y \log\left( \cos x \right)\]

Differentiating with respect to x using chain rule,

\[\frac{1}{y}\frac{dy}{dx} = y\frac{d}{dx}\left\{ \log \cos x \right\} + \log \cos x\frac{dy}{dx}\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = y\left( \frac{1}{\cos x} \right)\frac{d}{dx}\left( \cos x \right) + \log \cos x\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1}{y} - \log \cos x \right) = \frac{y}{\cos x}\left( - \sin x \right)\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1 - y \log \cos x}{y} \right) = - y \tan x\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{y^2 \tan x}{\left( 1 - y \log \cos x \right)}\]