# Implicit differentiation

• Dec 16th 2009, 04:35 PM
Da Freak
Implicit differentiation
could you tell me how to attack this one

http://img96.imageshack.us/img96/8578/dyoverdx.jpg
• Dec 16th 2009, 04:43 PM
skeeter
Quote:

Originally Posted by Da Freak
thanks a lot man.

could you tell me how to attack this one too
http://img96.imageshack.us/img96/8578/dyoverdx.jpg

implicit differentiation
• Dec 16th 2009, 04:46 PM
Da Freak
Quote:

Originally Posted by skeeter
implicit differentiation

how do you do that, i had a huge amount of trouble with that previously and got destroyed big time on that part of a test earlier in the year.
• Dec 16th 2009, 04:57 PM
skeeter
Quote:

Originally Posted by Da Freak
how do you do that, i had a huge amount of trouble with that previously and got destroyed big time on that part of a test earlier in the year.

$\displaystyle \frac{d}{dx} (x^2y^2+2y=x^3)$

$\displaystyle x^2 \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 2x + 2 \cdot \frac{dy}{dx} = 3x^2$

$\displaystyle 2x^2y \cdot \frac{dy}{dx} + 2 \cdot \frac{dy}{dx} = 3x^2 - 2xy^2$

$\displaystyle \frac{dy}{dx}(2x^2y + 2) = 3x^2 - 2xy^2$

$\displaystyle \frac{dy}{dx} = \frac{3x^2 - 2xy^2}{2x^2y + 2}$
• Dec 16th 2009, 05:50 PM
Da Freak
Quote:

Originally Posted by skeeter
$\displaystyle \frac{d}{dx} (x^2y^2+2y=x^3)$

$\displaystyle x^2 \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 2x + 2 \cdot \frac{dy}{dx} = 3x^2$

$\displaystyle 2x^2y \cdot \frac{dy}{dx} + 2 \cdot \frac{dy}{dx} = 3x^2 - 2xy^2$

$\displaystyle \frac{dy}{dx}(2x^2y + 2) = 3x^2 - 2xy^2$

$\displaystyle \frac{dy}{dx} = \frac{3x^2 - 2xy^2}{2x^2y + 2}$

thanks a lot man, i really appreciate the help.