# Thread: Find the second derivative of this ellipse

1. ## Find the second derivative of this ellipse

Find the second derivative of the ellipse
$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$

I know how to do the first derivative
it would be

$\frac{2x}{a^2} + \frac{2y}{b^2}(\frac{dy}{dx}) = 0$

but what comes next if I want to find the second derivative?

could someone show me full working out?

2. Originally Posted by differentiate
Find the second derivative of the ellipse
$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$

I know how to do the first derivative
it would be

$\frac{2x}{a^2} + \frac{2y}{b^2}(\frac{dy}{dx}) = 0$

but what comes next if I want to find the second derivative?

could someone show me full working out?

Solve for $\frac{dy}{dx}$.

I think you can see that

$\frac{2y}{b^2}\,\frac{dy}{dx} = -\frac{2x}{a^2}$

$\frac{dy}{dx} = -\frac{b^2x}{a^2y}$.

So $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{b^2x}{a^2y}\right)$

$= -\frac{d}{dx}\left(\frac{b^2x}{a^2y}\right)$

$= -\frac{a^2b^2y - a^2b^2x\frac{dy}{dx}}{a^4y^2}$

$= \frac{a^2b^2x\frac{dy}{dx} - a^2b^2y}{a^4y^2}$

$= \frac{b^2x\frac{dy}{dx} - b^2y}{a^2y^2}$.

Now substitute $\frac{dy}{dx}$.