$\displaystyle x^2+y^2=1$

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 $\displaystyle \frac{y}{x}$.

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- Nov 29th 2007, 10:05 PMTruthbetoldMore Implicit Differentiation
$\displaystyle x^2+y^2=1$

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 $\displaystyle \frac{y}{x}$. - Nov 29th 2007, 10:12 PMJhevon
- Nov 30th 2007, 04:40 AMkalagota
aww, i got $\displaystyle \frac{-1}{y^3}$..

- Nov 30th 2007, 05:09 AMcolby2152
$\displaystyle x^2 + y^2 = 1$

$\displaystyle 2x\frac{dy}{dx} + 2y = 0$

$\displaystyle \frac{dy}{dx} = \frac{-y}{x}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{-1}{x} + \frac{-y}{-x^2}\frac{dy}{dx} = \frac{-1}{x} + \frac{-y}{-x^2}\frac{-y}{x} = \frac{-1}{x} + \frac{-y^2}{x^3} = \frac{-(x^2+y^2)}{x^3}$

I got something different too. Hmm, did I miss something? - Nov 30th 2007, 05:41 PMkalagota
- Nov 30th 2007, 05:54 PMgalactus
- Nov 30th 2007, 06:38 PMcolby2152