# Partial defferentiation

1. Sep 11, 2008

### messedmonk

I'm not quite sure how this partial d/dx should look

(x + y) / sqr root(x^2 + y^2)

Is it (-2)(1/2x)(x+y)/((x^2 +y^2)^3/2) ??

2. Sep 11, 2008

### Defennder

Partial differentiation with respect to x is the same as doing differentiation with respect to x (ie. d/dx), only that every other variable such as y,z are considered constants. So, just replace y with a constant c and evaluate d/dx. When you're done, replace the c's with y and that's the answer.