# Implicit Differentation

• Dec 8th 2009, 03:20 PM
Crell
Implicit Differentation
I've been trying to figure it out, read quite a bit on it, but still struggling a little bit.

$2x^6+y^3-2x+2y=1062933$

and

$3125=(2y^4-4x^5)^5$

Here is what I worked out.

$2x^6+y^3-2x+2y=1062933$
$12x^5+(3y^2)y'-2+2y'=0$
$y'=\frac{-12x^5+2}{(3y^2)+2}$

I'm not too sure where to begin with the 2nd one. Also I'm pretty sure y' = dy/dx right?
• Dec 8th 2009, 03:37 PM
Quote:

Originally Posted by Crell
I've been trying to figure it out, read quite a bit on it, but still struggling a little bit.

$2x^6+y^3-2x+2y=1062933$

and

$3125=(2y^4-4x^5)^5$

Here is what I worked out.

$2x^6+y^3-2x+2y=1062933$
$12x^5+(3y^2)y'-2+2y'=0$
$y'=\frac{-12x^5+2}{(3y^2)+2}$

I'm not too sure where to begin with the 2nd one. Also I'm pretty sure y' = dy/dx right?

right

$3125=(2y^4-4x^5)^5$

For this one you have to use the chain rule. Think of it like this:

Let $u=2y^4-4x^5$

$0=\frac{d}{dx}(2y^4-4x^5)^5$

$=\left(\frac{d}{du}u^5\right)\left(\frac{du}{dx}\r ight)$

$=(5u^4)(6y^3y'-20x^4)$

$=5(2y^4-4x^5)^4(6y^3y'-20x^4)$

So now just solve for $y'$