Partial Derivative Question

**Gotovina7** · April 19th 2008, 02:37 PM

Can't seem to figure it out.

$\frac{d^{2}z}{dx^2}.......... z=arctan(\frac{y}{x})$

Thanks. I did not know how to make a tan inverse in Latex.

**TheEmptySet** · April 19th 2008, 03:03 PM

Originally Posted by Gotovina7

Can't seem to figure it out.

$\frac{d^{2}z}{dx^2}.......... z=arctan(\frac{y}{x})$

Thanks. I did not know how to make a tan inverse in Latex.

p.s the code is tan^{-1}

$\frac{\partial z}{\partial x}=\frac{1}{1+\left( \frac{y}{x} \right)^2}\cdot \left( -\frac{y}{x^2}\right)=\frac{-y}{x^2+y^2}$

taking one more gives

$\frac{\partial ^2 z}{\partial x^2}=\frac{2xy}{(x^2+y^2)^2}$

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