question_answer

If \[y={{(\cos \,x)}^{{{(\cos \,x)}^{(\cos \,x)...\,\infty }}}}\] , Show that \[\frac{dy}{dx}=\frac{{{y}^{2}}\tan x}{y\log \,\cos x-1}.\]

Answer:

We have, \[y={{(\cos \,x)}^{{{(\cos \,x)}^{(\cos \,x)...\,\infty }}}}\] On taking log both sides, we get             \[\log \,y=\log \,{{(\cos \,x)}^{{{(\cos \,x)}^{(\cos \,x)...\,\infty }}}}\] \[\Rightarrow \]   \[\log \,y=\log \,{{(\cos \,x)}^{y}}\] \[\Rightarrow \]   \[\log \,y=y\cdot \log \,\cos \,x\] \[[\because \,\,\log {{m}^{n}}=n\log m]\]On differentiating both sides w.r.t. x, we get \[\frac{1}{y}\cdot \frac{dy}{dx}=y\cdot \left[ \frac{1}{\cos x}(-\sin \,x) \right]+\log \,\cos x\cdot \frac{dy}{dx}\] \[\Rightarrow \]   \[\frac{1}{y}\cdot \frac{dy}{dx}=-\,y\,\tan \,x+\log \,\cos x\cdot \frac{dy}{dx}\]    \[\Rightarrow \]   \[\frac{1}{y}\cdot \frac{dy}{dx}-\log \,\cos x\cdot \frac{dy}{dx}=-\,y\tan x\] \[\Rightarrow \]   \[\frac{dy}{dx}\cdot \left( \frac{1-y\log \,\cos x}{y} \right)=-\,y\tan x\] \[\Rightarrow \]   \[\frac{dy}{dx}=\frac{-\,{{y}^{2}}\tan x}{1-y\log \,\cos x}\] \[\therefore \]      \[\frac{dy}{dx}=\frac{{{y}^{2}}\tan x}{y\log \,\cos x-1}\]         Hence proved.