Originally Posted by

**Hardwork** I've $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I'm asked to find $\displaystyle \frac{d^2y}{dx^2}$. I've implicitly differentiated it twice, but I can't get the required answer. What I did:

$\displaystyle y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{ dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2} +\frac{2y}{b^2}\right) = 0$

From this I got $\displaystyle \frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}$, whereas I was expected to get something quite different. So what has possibly gone wrong?