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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If y^ x =x^ y , then (d y)/(d x)=

If ycos⁡x=xsin⁡yy^{\cos x}=x^{\sin y}, then dydx=\frac{d y}{d x}=

Options

  1. Ay(xsin⁡xlog⁡y+sin⁡y)x(cos⁡x−ylog⁡xcos⁡y)\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}
  2. By(xsin⁡xlog⁡x−sin⁡y)x(cos⁡x+ylog⁡xcos⁡y)\frac{y(x \sin x \log x-\sin y)}{x(\cos x+y \log x \cos y)}
  3. Cy(sin⁡y−xlog⁡y)x(x−ycos⁡y(log⁡x))\frac{y(\sin y-x \log y)}{x(x-y \cos y(\log x))}
  4. Dy(sin⁡y+xlog⁡y)x(x+ycos⁡y(log⁡x))\frac{y(\sin y+x \log y)}{x(x+y \cos y(\log x))}

Correct answer

A. y(xsin⁡xlog⁡y+sin⁡y)x(cos⁡x−ylog⁡xcos⁡y)\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}

Step-by-step solution

It is given that ycos⁡x=xsin⁡yy^{\cos x}=x^{\sin y} On taking logrithm both sides, we get cos⁡xlog⁡y=sin⁡ylog⁡x\cos x \log y=\sin y \log x On differentiating both side w.r.t. xx, we get 1y(cos⁡x)dydx−(sin⁡x)log⁡y=1xsin⁡y+(log⁡x)(cos⁡y)ddx\frac{1}{y}(\cos x) \frac{d y}{d x}-(\sin x) \log y=\frac{1}{x} \sin y+(\log x)(\cos y) \frac{d}{d x}⇒dydx(1ycos⁡x−(log⁡x)cos⁡y)=1xsin⁡y+(sin⁡x)log⁡y⇒dydx=y(sin⁡y+xsin⁡xlog⁡y)x(cos⁡x−ylog⁡xcos⁡y)\begin{aligned} & \Rightarrow \frac{d y}{d x}\left(\frac{1}{y} \cos x-(\log x) \cos y\right)=\frac{1}{x} \sin y+(\sin x) \log y \\ & \Rightarrow \frac{d y}{d x}=\frac{y(\sin y+x \sin x \log y)}{x(\cos x-y \log x \cos y)}\end{aligned}$$
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