Question

Differentiate $y={\sin x}^{{\sin x}^{{\sin x}^{}....\infty }}$

• A. $\frac{dy}{dx}=\frac{y^2\cot x}{(1-y\log \sinh x)}$
• B. $\frac{dy}{dx}=\frac{y^2\cot x}{(1-y\log \cos x)}$
• C. $\frac{dy}{dx}=\frac{y^2\cot x}{(1-y\log \sin x)}$
• D. None of these

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{y^2\cot x}{(1-y\log \sin x)}$

$y={\sin x}^{{\sin x}^{...\infty }}$

$y={\sin x}^y$

$\ln y=y\ln x\sin x$

Differentiating wrt $x$

$\ln y=y\ln x$

$\Rightarrow \frac{y^{\prime }}{y}=\frac{y\cos x}{x\sin x}+y^{\prime }\ln x$

$y^{\prime }=\frac{y^2}{x\tan x}+y^{\prime }y\ln x\sin x$

$y^{\prime }=\frac{\frac{y^2}{x\tan x}}{(1-y)\ln x}=\frac{\frac{y^2}{\tan x}}{1-y\ln x\sin x}=\frac{y^2\cot x}{1-y\ln x\sin x}$