$\displaystyle x^2+y^2=1$
Find dy/dx and the second derivative.
The answer says both derivatives equal $\displaystyle \frac{-x}{y}$.
The first one does, but the second derivative I got as $\displaystyle \frac{y}{x}$.
$\displaystyle x^2 + y^2 = 1$
$\displaystyle 2x\frac{dy}{dx} + 2y = 0$
$\displaystyle \frac{dy}{dx} = \frac{-y}{x}$
$\displaystyle \frac{d^2y}{dx^2} = \frac{-1}{x} + \frac{-y}{-x^2}\frac{dy}{dx} = \frac{-1}{x} + \frac{-y}{-x^2}\frac{-y}{x} = \frac{-1}{x} + \frac{-y^2}{x^3} = \frac{-(x^2+y^2)}{x^3}$
I got something different too. Hmm, did I miss something?