How do I find dy/dx of this problem?

$\displaystyle tan (x/y) = x + y$

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- Feb 25th 2008, 10:18 PM!!!Finding dy/dx by implicit differentiation
How do I find dy/dx of this problem?

$\displaystyle tan (x/y) = x + y$ - Feb 25th 2008, 10:35 PMJhevon
do you realize that everytime you differentiate a y-term you attach dy/dx to it?

so, to start you off.

$\displaystyle \tan \frac xy = x + y$

$\displaystyle \Rightarrow \sec^2 \left( \frac xy \right) \cdot \frac d{dx} \left( \frac xy \right) = 1 + \frac {dy}{dx}$

(the left hand side follows by the chain rule)

can you continue? - Feb 26th 2008, 01:19 AM!!!
I actually forgot what I did, redid the problem and got the answer correct, but thank you!

Can you help me with another one, please?

$\displaystyle xy = cot (xy)$ - Feb 26th 2008, 01:38 AMJhevon