Partial Derivatives with Inverse Trig Functions

[issisoccer10]

[SOLVED] Partial Derivatives with Inverse Trig Functions

Homework Statement

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

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}

given that u = ln(x^{2} + y^{2})
and that v = 2tan^{-1} (y/x)

Homework Equations

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

\frac{d}{dx} ln (u) = \frac{1}{u} \frac{du}{dx}

\frac{d}{dx} tan^{-1} u = \frac{1}{1+u^{2}} \frac{du}{dx}

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 \frac{du}{dx} there would be \frac{\partial u}{\partial x} or \frac{\partial u}{\partial y} depending on which the question called for. With this reasoning, I thought the left side of the Cauchy-Riemann equation should be..

\frac{2x}{x^{2} + y ^{2}}

and the right side would be...

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

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

 

[NateTG]

\frac{2x}{x^{2} + y ^{2}}

and the right side would be...

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

\frac{2}{1+\left(\frac{y}{x}\right)^2} \times \frac{1}{x}
Distribute through and multply by 1 (\frac{x}{x])...
\frac{2}{x+\frac{y^2}{x}} \times \frac{x}{x}
\frac{2x}{x^2+y^2}

 

[issisoccer10]

in the denominator of your responce you have the term y^{2}/x. Shouldn't the term be y^{2}/x^{2}?

 

[NateTG]

\frac{1}{1+\left(\frac{y}{x}\right)^2} \times \frac{1}{x}
\frac{1}{\left(1+\left(\frac{y}{x}\right)^2\right) x}
\frac{1}{x + x \times \frac{y^2}{x^2}}
\frac{1}{x + \frac{y^2}{x}}

 

[issisoccer10]

wow... i feel pathetic... thanks again.