If y=(sin⁡x)(sin⁡x)(sin⁡x).....∞, then dydx=

y=(sin⁡x)y⇒log⁡y=y⋅log⁡(sin⁡x)1y⋅dydx=y⋅cot⁡x+log⁡(sin⁡x)dydx⇒dydx1y−log⁡(sin⁡x)=ycot⁡x⇒dydx1−y⋅log⁡(sin⁡x)y=ycot⁡x⇒dydx=y2cot⁡x1−y⋅log⁡(sin⁡x)=y2⋅cot⁡x1−log⁡y.

If y=(sin⁡x)(sin⁡x)(sin⁡x).....∞, then dydx=

y=(sin⁡x)y⇒log⁡y=y⋅log⁡(sin⁡x)1y⋅dydx=y⋅cot⁡x+log⁡(sin⁡x)dydx⇒dydx1y−log⁡(sin⁡x)=ycot⁡x⇒dydx1−y⋅log⁡(sin⁡x)y=ycot⁡x⇒dydx=y2cot⁡x1−y⋅log⁡(sin⁡x)=y2⋅cot⁡x1−log⁡y.

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If $y = (\sin x)^{(\sin x)^{{(\sin x)}^{. . . . . \infty}}}$ , then $\frac{dy}{dx} =$

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$\frac{y^{2}}{\sin x [1 - \log y]}$

$\frac{y^{2} \sin x}{1 - \log y}$

$\frac{y^{2} \cot x}{1 - \log y}$

$\frac{y^{2} \tan x}{1 - \log y}$

answer is C.

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Detailed Solution

$\begin{array}{l} y = (\sin x)^{y} \Rightarrow \log y = y \cdot \log (\sin x) \\ \frac{1}{y} \cdot \frac{dy}{dx} = y \cdot \cot x + \log (\sin x) \frac{dy}{dx} \Rightarrow \frac{dy}{dx} [\frac{1}{y} - \log (\sin x)] = ycot x \\ \Rightarrow \frac{dy}{dx} [\frac{1 - y \cdot \log (\sin x)}{y}] = ycot x \Rightarrow \frac{dy}{dx} = \frac{y^{2} \cot x}{1 - y \cdot \log (\sin x)} = \frac{y^{2} \cdot \cot x}{1 - \log y} . \end{array}$