Finding dy/dx by implicit differentiation

**!!!** · February 25th 2008, 10:18 PM

How do I find dy/dx of this problem?

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

**Jhevon** · February 25th 2008, 10:35 PM

Originally Posted by !!!

How do I find dy/dx of this problem?

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

do you realize that everytime you differentiate a y-term you attach dy/dx to it?

so, to start you off.

$\tan \frac xy = x + y$

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

(the left hand side follows by the chain rule)

can you continue?

**!!!** · February 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?
$xy = cot (xy)$

**Jhevon** · February 26th 2008, 01:38 AM

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?
$xy = cot (xy)$

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

to start you off.

$xy = \cot (xy)$

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

can you continue?

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