- #1
Warr
- 120
- 0
[tex]e^{xy} = ln (x+y)[/tex]
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?
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?