- #1

issisoccer10

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**[SOLVED] Partial Derivatives with Inverse Trig Functions**

## Homework Statement

Show that u(x,y) and v(x,y) satisfy the Cauchy-Riemann equations...

[tex]\frac{\partial u}{\partial x}[/tex] = [tex]\frac{\partial v}{\partial y}[/tex]

given that u = ln(x[tex]^{2}[/tex] + y[tex]^{2}[/tex])

and that v = 2tan[tex]^{-1}[/tex] (y/x)

## Homework Equations

single variable differentials of equations should be in the form of..

[tex]\frac{d}{dx}[/tex] ln (u) = [tex]\frac{1}{u}[/tex] [tex]\frac{du}{dx}[/tex]

[tex]\frac{d}{dx}[/tex] tan[tex]^{-1}[/tex] u = [tex]\frac{1}{1+u^{2}}[/tex] [tex]\frac{du}{dx}[/tex]

## The Attempt at a Solution

I would think that the partial derivatives of the above equations would lead from the single differentials, but instead of [tex]\frac{du}{dx}[/tex] there would be [tex]\frac{\partial u}{\partial x}[/tex] or [tex]\frac{\partial u}{\partial y}[/tex] depending on which the question called for. With this reasoning, I thought the left side of the Cauchy-Riemann equation should be..

[tex]\frac{2x}{x^{2} + y ^{2}}[/tex]

and the right side would be...

[tex]\frac{2}{1 + (y/x)^{2}}[/tex] [tex]\frac{1}{x}[/tex]

However, these don't appear to be equal... unless I can't see the simple algebra. Any help would be greatly appreciated..