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enter image description here I have done the sum by taking log but not getting my final answer right. Please help me

Andrew Chin
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L Lawlit
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  • Why do you believe that answer is incorrect? Keep in mind that, because of the known relation between $x$ and $y$, there are likely to be many equivalent ways to write the solution. Also, you should simplify your fraction. – lulu Nov 07 '19 at 17:36

2 Answers2

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Given $x^y=e^{x-y}$, find $\dfrac{dy}{dx}$.

\begin{align} x^y&=e^{x-y}\\ \ln (x^y)&=x-y\\ y\ln x&=x-y\\ \frac{d}{dx}(y\ln x)&=\frac{d}{dx}(x-y)\\ \frac{dy}{dx}\ln x+\frac{y}{x}&=1-\frac{dy}{dx}\\ (1+\ln x)\frac{dy}{dx}&=1-\frac{y}{x}\\ \frac{dy}{dx}&=\frac{1-\frac{y}{x}}{1+\ln x}\\ \frac{dy}{dx}&=\boxed{\frac{x-y}{x+x\ln x}} \end{align}

Edit to match provided answer: \begin{align} x^y&=e^{x-y}\\ \log (x^y)&=x-y\\ y\log x&=x-y\\ y+y\log x&=x\\ y(1+\log x)&=x\\ y&=\frac{x}{1+\log x}\\ \frac{dy}{dx}&=\frac{(1+\log x)-x(\frac1x)}{(1+\log x)^2}\\ \frac{dy}{dx}&=\boxed{\frac{\log x}{(1+\log x)^2}} \end{align}

Andrew Chin
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  • Actually the correct answer is given as log x/(1+log x) ² – L Lawlit Nov 07 '19 at 17:45
  • @KalpitPrabhu compare now. – Andrew Chin Nov 07 '19 at 17:50
  • For the future, you may be able to get the most appropriate answer by outlining the topic at hand (obviously from the above, you can see that the derivative $\frac{dy}{dx}$ can be achieved by both differentiating an explicit function $y$ in terms of $x$ OR by implicit differentiation). – Andrew Chin Nov 07 '19 at 17:52
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It is $$y\ln(x)=x-y$$ differentiating with respect to $x$ we get $$y'\ln(x)+\frac{y}{x}=1-y'$$ so we obtain $$y'(\ln(x)+1)=\frac{x-y}{x}$$ or $$y'=\frac{x-y}{x(\ln(x)+1)}$$

Dr. Sonnhard Graubner
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