1. ## Implicit Differentation

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

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

and

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

Here is what I worked out.

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

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

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

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

and

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

Here is what I worked out.

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

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

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

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

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

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

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

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

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

So now just solve for $\displaystyle y'$