Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

$$x^y = y^x$$

find an expression for $dy/dx$ in terms of $y$ and $x$

for the first part is $yx^{y-1}$ and for the second part i got $\ln(y)y^xdy/dx$

when solving for dy/dx i got $yx^{y-1}) / (\ln(y)y^x)$

Just was wondering if my work is correct?

Sorry for the bad formatting, Im not sure how to make it formatted nice.

share|cite|improve this question
what do you do to get the formatting of the exponents? – user71317 Apr 6 '13 at 23:02
up vote 3 down vote accepted

It might be easier to take $\log$ first and work with the expression. We have $$y \log x = x \log y$$ Now differentiating we get $$y' \log x +\dfrac{y}x = \log y + x \dfrac{y'}y \implies y' \left(\log x - \dfrac{x}y \right) = \left(\log y - \dfrac{y}x \right)$$ Hence, we get that $$y' = \dfrac{\left(\log y - \dfrac{y}x \right)}{\left(\log x - \dfrac{x}y \right)}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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