what is the second derivative of x^2 + y^2 = s^2
thanks
Ummm... You are looking for $\displaystyle \frac{d^2y}{dx^2}$? And I presume s is a constant?
$\displaystyle 2x + 2y \frac{dy}{dx} = 0$
$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$
$\displaystyle \frac{d^2y}{dx^2} = -\frac{y - x \frac{dy}{dx}}{y^2}$
Insert the equation for $\displaystyle \frac{dy}{dx}$:
$\displaystyle \frac{d^2y}{dx^2} = -\frac{y - x \cdot -\frac{x}{y}}{y^2}$
$\displaystyle \frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3}$
Inserting the original equation:
$\displaystyle \frac{d^2y}{dx^2} = -\frac{s^2}{y^3}$
I'd probably leave it like that.
-Dan