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Mathematics Question on limits and derivatives

If y=(sin⁡x)(sin⁡x)(sin⁡x)....∞,y = \left(\sin x\right)^{\left(\sin x\right)^{\left(\sin x\right)....\infty}} , then dydx=\frac{dy}{dx} =

A

y2sin⁡x(1−log⁡y)\frac{y^{2}}{\sin x\left(1-\log y\right)}

B

y2sin⁡x1−log⁡y\frac{y^{2} \sin x}{1-\log y}

C

y2cot⁡x1−log⁡y\frac{y^{2} \cot x}{1-\log y}

D

y2tan⁡x1−log⁡y\frac{y^{2} \tan x}{1-\log y}

Answer

y2cot⁡x1−log⁡y\frac{y^{2} \cot x}{1-\log y}

Explanation

Solution

y=(sin⁡x)y⇒log⁡y=ylog⁡sin⁡xy = \left(\sin x\right)^{y} \Rightarrow \log y = y \log\sin x ⇒1ydydx=dydxlog⁡(sin⁡x)+y1sin⁡x.cos⁡x \Rightarrow \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log\left(\sin x\right) + y \frac{1}{\sin x} . \cos x ⇒dydx(1y−log⁡sin⁡x)=ycot⁡x\Rightarrow \frac{dy}{dx} \left( \frac{1}{y} - \log\sin x\right) = y \cot x ⇒dydx=y2cot⁡x1−ylog⁡sin⁡x=y2cot⁡x1−log⁡y[∵log⁡y=log⁡sin⁡x]\Rightarrow \frac{dy}{dx} = \frac{y^{2} \cot x }{1 -y \log\sin x} = \frac{y^{2} \cot x}{1-\log y} \left[\because\log y = \log\sin x \right]