How to Apply Partial Differentiation to V=f(x²+y²)?

[patrickmoloney]

Homework Statement

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

Homework Equations

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.

 

[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. $$\partial_x V = \frac{df}{dg}\partial_x g(x,y)$$

 

[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

 

[Simon Bridge]

Well done :)

 

[HalfLostAndSearching]

I would suggest starting by rewriting the function V in terms of x and y, instead of f(x) and f(y). This will make it easier to apply partial differentiation. So, let's start with V = f(x²+y²) = f(u), where u = x²+y².

Now, we can use the chain rule to find ∂V/∂x and ∂V/∂y:

∂V/∂x = ∂f/∂u * ∂u/∂x = 2xf'(u) = 2xf'(x²+y²)

∂V/∂y = ∂f/∂u * ∂u/∂y = 2yf'(u) = 2yf'(x²+y²)

Next, we can substitute these expressions into the given equation:

x(∂V/∂y) - y(∂V/∂x) = x * 2yf'(x²+y²) - y * 2xf'(x²+y²) = 2xyf'(x²+y²) - 2xyf'(x²+y²) = 0

Therefore, we have shown that x(∂V/∂y) - y(∂V/∂x) = 0, as required. This is a useful result in partial differentiation, and it can also be extended to functions of more than two variables. I hope this helps!