Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $y'$?

share|cite|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas, Edward Jiang, Luiz Cordeiro, Giovanni, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.