Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x})^{...\infty}}}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to
Options:
Solution:
2384 Upvotes
Verified Answer
The correct answer is:
$-\frac{y^{2} \tan x}{1-y \operatorname{In}(\cos x)}$
$\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x})^{...\infty}}}$
$\Rightarrow \mathrm{y}=(\cos \mathrm{x})^{\mathrm{y}}$
$\Rightarrow \log \mathrm{y}=\mathrm{y} \cdot \log \cos \mathrm{x}$
Differentiating on both sides, we get
$\frac{1}{y} \cdot \frac{d y}{d x}=y(-\tan x)+\log \cos x \cdot \frac{d y}{d x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{1}{\mathrm{y}}-\log \cos \mathrm{x}\right)=-\mathrm{y} \tan \mathrm{x}$
$\Rightarrow \frac{d y}{d x}=\frac{-y \tan x}{\frac{1}{y}-\log \cos x}=\frac{-y^{2} \tan x}{1-y \log \cos x}$
$\Rightarrow \mathrm{y}=(\cos \mathrm{x})^{\mathrm{y}}$
$\Rightarrow \log \mathrm{y}=\mathrm{y} \cdot \log \cos \mathrm{x}$
Differentiating on both sides, we get
$\frac{1}{y} \cdot \frac{d y}{d x}=y(-\tan x)+\log \cos x \cdot \frac{d y}{d x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{1}{\mathrm{y}}-\log \cos \mathrm{x}\right)=-\mathrm{y} \tan \mathrm{x}$
$\Rightarrow \frac{d y}{d x}=\frac{-y \tan x}{\frac{1}{y}-\log \cos x}=\frac{-y^{2} \tan x}{1-y \log \cos x}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.