Hi guys, having some trouble on implicit differentiation. The problem is
find y'' given x^2 - y^2 = 1
I got the answer (y^2-x^2)/y^3
Remember that you can treat y's just like the chain rule with x's. So:
$\displaystyle \frac{d}{dx}y^2=2y(y')$
and so on. Taking our first derivative:
$\displaystyle \frac{d}{dx}[y^2-x^2]=\frac{d}{dx}1$
$\displaystyle 2yy'-2x=0$
$\displaystyle \frac{d}{dx}[2yy'-2x]=\frac{d}{dx}0$
$\displaystyle 2yy''+2(y')^2-2=0$
To get the answer you had posted, you just substitute from above:
$\displaystyle 2yy'-2x=0$
$\displaystyle y'=\frac{x}{y}$
and you get:
$\displaystyle 2yy''+2(\frac{x}{y})^2-2=0$
which can be rearranged thusly:
$\displaystyle y''=\frac{y^2-x^2}{y^3}$