• May 5th 2008, 03:33 AM
4sweet2005
• May 5th 2008, 03:42 AM
Moo
Hello,

$f(x,y)=\sqrt{\frac{x+y}{x-y}}$

The gradient vector of a function is defined as following :

$\vec{\nabla}f=\left(\frac{\partial f}{\partial x} \ , \ \frac{\partial f}{\partial y} \right)$

For example, $\frac{\partial f}{\partial x}$ is the derivative of f, considering that x is the variable and y is a constant.
• May 5th 2008, 03:47 AM
4sweet2005
This is what I just worked out ( 2y/(x-y)^2 , -2y/(x-y)^2 ), not sure if that's the answer?
• May 5th 2008, 03:54 AM
topsquark
Quote:

Originally Posted by Moo
$f(x,y)=\sqrt{\frac{x+y}{x-y}}$

Quote:

Originally Posted by 4sweet2005
This is what I just worked out ( 2y/(x-y)^2 , -2y/(x-y)^2 ), not sure if that's the answer?
$\frac{\partial }{\partial x} \left ( \sqrt{\frac{x+y}{x-y}} \right )$
$= \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x+y}{x-y}}} \cdot \left ( \frac{(1)(x - y) - (x + y)(1)}{(x - y)^2} \right )$
$= \sqrt{\frac{x-y}{x+y}} \cdot \left ( \frac{-2y}{2(x - y)^2} \right )$
$= -\frac{y}{(x - y)^{3/2} (x + y)^{1/2}}$