# Partial differentiation

1. Dec 14, 2013

### patrickmoloney

1. The problem statement, all variables and given/known data

let V=f(x²+y²) , show that x(∂V/∂y) - y(∂V/∂x) = 0

2. Relevant equations

3. The attempt at a solution

V=f(x²+y²) ; V=f(x)² + f(y)²

∂V/∂x = 2[f(x)]f'(x) + [0]

∂V/∂y = 2[f(y)]f'(y)

I'm sure I've gone wrong somewhere, I have never seen functions like this, I'm just used to using V=f(x,y)= some function and then partially differentiations. help would be much apprectiated.

2. Dec 14, 2013

### Simon Bridge

You are nearly there:
You have a function of form: $V(x,y)=f(g(x,y))$
... so you'd use the chain rule. $$\partial_x V = \frac{df}{dg}\partial_x g(x,y)$$

3. Dec 14, 2013

### patrickmoloney

V(x,y) = f(g(x,y)

using chain rule:

∂V/∂x = df/dg (∂g/∂x[(x²+y²)])
= 2x(df/dg)

y(∂V/∂x) = 2xy(df/dg)

∂V/∂y = df/dg (∂g/∂y[(x²+y²)])
= 2y(df/dg)

x(∂V/∂y) = 2xy(df/dg)

x(∂V/∂y)-y(∂V/∂x) = 0

2xy(df/dg) - 2xy(df/dg) = 0

4. Dec 14, 2013

Well done :)