[tex]e^{xy} = ln (x+y)[/tex](adsbygoogle = window.adsbygoogle || []).push({});

I need to find dy/dx...but its difficult to get the answer in the book

I tried this:

[tex]ln e^{xy} = ln (ln (x + y))[/tex]

[tex]xy = ln (ln (x+y))[/tex]

taking the derivitive in terms of x

[tex]y + \frac{dy}{dx}x = (\frac {1}{ln (x+y)})(\frac {1}{x+y})(1 + \frac {dy}{dx})[/tex]

If I were to continue and solve for dy/dx It would not even be close to the books answer. I'm probably doing it wrong anyways...can someone show me the right way?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Derivative Problem

**Physics Forums | Science Articles, Homework Help, Discussion**