If y^x=e^{y-x}, then \;d y/d x is equal to

Correct Answer is : (1+log⁡y)2log⁡y\;(1+ y)²/ ylogy(1+logy)2

Correct Answer is : (1+log⁡y)2log⁡y\;(1+ y)²/ ylogy(1+logy)2

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BITSAT 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 111 of 130

Marks: +1, -0

y^x=e^{y-x}

\;\frac{d y}{d x}

Solution: 👈: Video Solution

Here,

y^{x}=e^{y-x}

Taking

\log

on both sides, we get

\;\log y^{x}=\log e^{y-x}

\;\;\;\;(\because \log a^{b}=b\log a\;\text{ and }\;\log e=1)

\Rightarrow x\log y=(y-x)\log e\Rightarrow x\log y=y-x

On differentiating w.r.t.

x

, we get

\;\frac{d}{dx}(x\log y)=\;\frac{d}{dx}(y-x)

(using product rule)

\;\Rightarrow x\left(\frac{1}{y}\right)\;\frac{d y}{d x}+\log y(1)=\;\frac{d y}{d x}-1

\;\Rightarrow\;\frac{d y}{d x}\left(\frac{x}{y}-1\right)=-1-\log y

\;\Rightarrow\;\frac{d y}{d x}\left[\frac{y}{(1+\log y)y}-1\right]=-(1+\log y)

\;\Rightarrow\;\frac{d y}{d x}\left[\frac{1-1-\log y}{1+\log y}\right]=-(1+\log y)

\;\Rightarrow\;\frac{d y}{d x}=-\frac{(1+\log y)^2}{-\log y}

\Rightarrow\;\frac{d y}{d x}=\;\frac{(1+\log y)^2}{\log y}

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