Partial Derivative of atan(xy/(1+x^2+y^2)^0.5)

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SUMMARY

The discussion centers on proving the second mixed partial derivative of the function \( z = \arctan\left(\frac{xy}{\sqrt{1+x^2+y^2}}\right) \). The conclusion reached is that \( \frac{\partial^2 z}{\partial x \partial y} = \frac{1}{(1+x^2+y^2)^{\frac{3}{2}}} \). The initial differentiation with respect to \( y \) yielded \( \frac{\partial z}{\partial y} = \frac{x^3+x}{(1+x^2+y^2)^{\frac{3}{2}}(1+x^2+y^2+x^2y^2)} \), which was confirmed to be correct before proceeding to the second derivative.

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MrWarlock616
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Homework Statement



Prove that if ##z=\arctan(\frac{xy}{\sqrt(1+x^2+y^2)})## , then:

##\frac{\partial^2 z}{\partial x \partial y}=\frac{1}{(1+x^2+y^2)^\frac{3}{2}} ##

Homework Equations



##\frac{d}{d x} (\arctan(x)) = \frac{1}{1+x^2}##

The Attempt at a Solution



Differentiating z partially w.r.t y, I got:

## \frac{\partial z}{\partial y} = \frac{x^3+x}{(1+x^2+y^2)^\frac{3}{2} (1+x^2+y^2+x^2y^2)} ##

I'm pretty sure my working till here is correct. But differentiating again w.r.t x would be disastrous. Help me please. :)
 
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Hi MrWarlock616! :smile:

You can cancel a (1 + x2), top-and-bottom :wink:
 
EDIT: -nvm got it-
 

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