# Finding dy/dx by implicit differentiation

• 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 PM
Jhevon
Quote:

Originally Posted by !!!
How do I find dy/dx of this problem?

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

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 AM
Jhevon
Quote:

Originally Posted by !!!
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?
$\displaystyle xy = cot (xy)$

use the product rule on the left and the chain rule (coupled with the product rule) on the right

to start you off.

$\displaystyle xy = \cot (xy)$

$\displaystyle \Rightarrow y + x~\frac {dy}{dx} = - \csc^2 (xy) \cdot \frac d{dx} [xy]$

can you continue?