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.