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I have to find the derivative of $y=e^{-x/y}$..should I do this by taking the $\ln$ of both sides? Will that give me $y'$?

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closed as off-topic by Jonas, Edward Jiang, Luiz Cordeiro, Giovanni, choco_addicted May 4 at 2:16



It would be useful if you could show us your effort... – Nameless Dec 4 '12 at 19:44

Method 1:Try to take the derivative with respect to $x$ on both sides (use implicite differentiation): $$ \dfrac{dy}{dx} = e^{-x/y}\frac{d}{dx}(-x)y^{-1}. $$ You would get $$ \dfrac{dy}{dx} = -e^{-x/y}\left[y^{-1} - xy^{-2}\frac{dy}{dx}\right]. $$ Now try to solve this for $\displaystyle{\frac{dy}{dx}}$.

Method 2: You can indeed also first take a $\ln$ on both sides so that you get: $$ \ln(y) = -\frac{x}{y} = -xy^{-1}. $$ Again, take $\displaystyle{\frac{d}{dx}}$ on both sides and get $$ \frac{1}{y}\frac{dy}{dx} = -y^{-1}+xy^{-2}\frac{dy}{dx}. $$ Using that $y = e^{-x/y}$ these two methods actually give the same answer.

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