Find the gradient of $\frac{x}{x-y}$

Question

It seems simple on the face of it, but I cannot figure out how to actually do this. I know that you have to find the partial with respect to $x$ and also with respect to $y$, but that's where I get lost.

Answer 1

$$\left.\begin{array}{rcl}\frac{\partial}{\partial x}\left( \frac{x}{x-y}\right) = \frac{(x-y)-x}{(x-y)^2} &=&\frac{-y}{(x-y)^2}\\ \frac{\partial}{\partial y} \left(\frac{x}{x-y}\right) &=& \frac{x}{(x-y)^2}\end{array}\right\} \Longrightarrow \nabla \left(\frac{x}{x-y}\right) = \frac{1}{(x-y)^2}\begin{pmatrix}- y \\ x\end{pmatrix}$$

Both times, you should use the derivation rule: $$\frac{d}{dz}\left(\frac{f(z)}{g(z)}\right) = \frac{f'(z)g(z)-g'(z)f(z)}{g^2(z)},$$ one time with $f(z) = z, g(z) = z-y$ and the other time with $f(z) = 1, g(z) = x-z$

Answer 2

$f(x,y)=x/(x-y)$

Just use the definition

$\bigtriangledown f(x,y)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$

$\frac{\partial f}{\partial x}=-y/(x-y)^2$

$\frac{\partial f}{\partial y}=x/(x-y)^2$

so

$\bigtriangledown f(x,y)=(-y/(x-y)^2,x/(x-y)^2)$