# Thread: take the second derivative..

1. ## take the second derivative..

take the second derivative of x^2y^2-2x=3

2. Originally Posted by thatonegurl
take the second derivative of x^2y^2-2x=3
show us the first derivative ... you need that to get the second.

3. I'm confused with derivatives all together.

As far as the first derivative, I got as far as 2x2yy'-2=0

I'm not sure what to do after that, and I don't even know if that is right.

4. Originally Posted by thatonegurl
I'm confused with derivatives all together.

As far as the first derivative, I got as far as 2x2yy'-2=0

I'm not sure what to do after that, and I don't even know if that is right.
you need to use the product rule for the first term ...

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

$x^2 \cdot 2y \cdot y' + y^2 \cdot 2x - 2 = 0$

solve for $y'$ , then take the derivative again.