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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Differentiation

NDA 2017 · Mathematics · Differentiation

If y=( x)^ ( x)^ ( x)^ (∞)/(x) , then dy dx is equal to

If y=(cos⁡x)(cos⁡x)(cos⁡x)∞x\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos x)^{\frac{\infty}{x}}}}, then dydx\frac{\mathrm{dy}}{\mathrm{dx}} is equal to

Options

  1. A−y2tan⁡x1−yIn⁡(cos⁡x)-\frac{y^2 \tan x}{1-y \operatorname{In}(\cos x)}
  2. By2tan⁡x1+yIn⁡(cos⁡x)\frac{y^2 \tan x}{1+y \operatorname{In}(\cos x)}
  3. Cy2tan⁡x1−yIn⁡(sin⁡x)\frac{y^2 \tan x}{1-y \operatorname{In}(\sin \mathrm{x})}
  4. Dy2sin⁡x1+yIn⁡(sin⁡x)\frac{y^2 \sin x}{1+y \operatorname{In}(\sin x)}

Correct answer

A. −y2tan⁡x1−yIn⁡(cos⁡x)-\frac{y^2 \tan x}{1-y \operatorname{In}(\cos x)}

Step-by-step solution

y=(cos⁡x)(cos⁡x)cos⁡x⋯⇒y=(cos⁡x)y\begin{aligned} & y=(\cos x)^{(\cos x)^{\cos x \cdots}} \\ & \Rightarrow y=(\cos x)^y\end{aligned}\Rightarrow \log y=y \cdot \log \cos xDifferentiatingonbothsides,wegetDifferentiating on both sides, we get1y⋅dydx=y(−tan⁡x)+log⁡cos⁡x⋅dydx⇒dydx(1y−log⁡cos⁡x)=−ytan⁡x\begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=y(-\tan x)+\log \cos x \cdot \frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}\left(\frac{1}{y}-\log \cos x\right)=-y \tan x\end{aligned}⇒dydx=−ytan⁡x1y−log⁡cos⁡x=−y2tan⁡x1−ylog⁡cos⁡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}
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