# Thread: y' and y'' using implicit differentiation

1. ## y' and y'' using implicit differentiation

Hi guys, having some trouble on implicit differentiation. The problem is

find y'' given x^2 - y^2 = 1

2. Originally Posted by sleepiiee
Hi guys, having some trouble on implicit differentiation. The problem is

find y'' given x^2 - y^2 = 1

Remember that you can treat y's just like the chain rule with x's. So:

$\frac{d}{dx}y^2=2y(y')$

and so on. Taking our first derivative:

$\frac{d}{dx}[y^2-x^2]=\frac{d}{dx}1$

$2yy'-2x=0$

$\frac{d}{dx}[2yy'-2x]=\frac{d}{dx}0$

$2yy''+2(y')^2-2=0$

To get the answer you had posted, you just substitute from above:

$2yy'-2x=0$

$y'=\frac{x}{y}$

and you get:

$2yy''+2(\frac{x}{y})^2-2=0$

which can be rearranged thusly:

$y''=\frac{y^2-x^2}{y^3}$

3. thanks alot