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  • If yx = ey-x, prove that dy/dx=1+logy^2

If y^x = e^{y-x}, prove that \frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}

Answers (1)

Given:.

y^x = e^{y-x}

Taking log on both the sides, we get

\begin{aligned} &\log y^x=\log e^{y-x} \\ &\Rightarrow x \log y=y-x\\ \end{aligned}

\log y=\frac{y}{x}-1

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

\\ \Rightarrow \frac{d y}{d x}=\frac{\log y+1}{\frac{y-x}{y}} \\ \Rightarrow \frac{d y}{d x}=\frac{y[\log y+1]}{y-x} \\ \Rightarrow \frac{d y}{d x}=\frac{y[\log y+1]}{x \log y} \\ \Rightarrow \frac{d y}{d x}=\frac{[\log y+1] (\log y+1)}{\log y} \quad[u \sin g(i)] \\ \Rightarrow \frac{d y}{d x}=\frac{(\log y+1)^{2}}{\log y}

Hence Proved

Posted by

infoexpert22

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