Browse Questions

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

Toolbox:
• A function $f(x,y)$ is said to be implicit if it is jumbled in such a way,that it is not possible to write $y$ exclusively s a function of $x$.
• $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
Step 1:
Given : $y=(\cos x)^{\Large(\cos x)^{(\Large \cos x)..........\infty.}}$
$\Rightarrow y=(\cos x)^y$
Take $log$ on both sides,
$\log y=y\log \cos x$
Differentiate w.r.t $x$ on both sides,(Apply product rule to differentiate $y\cos x$
$\large\frac{1}{y}\frac{dy}{dx}=y.\large\frac{1}{\cos x}.$$-\sin x+\cos x.\large\frac{dy}{dx} \large\frac{1}{y}\frac{dy}{dx}$$=-y\tan x+\cos x\large\frac{dy}{dx}$
Step 2:
$\large\frac{dy}{dx}$$(\large\frac{1}{y}$$-\log(\cos x))=-y\sin x$
$\large\frac{dy}{dx}=\large\frac{-y^2\tan x}{1-y\log(\cos x)}$
$\Rightarrow \large\frac{y^2\tan x}{y\log (\cos x)-1}$
Therefore $\large\frac{dy}{dx}=\large\frac{y^2\tan x}{y\log (\cos x)-1}$
Hence proved.