# Homogeneity and derivatives

1. Nov 3, 2014

### Sidney

I've been reading a book on economics and they defined a homogeneous function as: ƒ(x1,x2,…,xn) such that
ƒ(tx1,tx2,…,txn)=tkƒ(x1,x2,…,xn) ..totally understandable.. they further explained that a direct result from this is that the partial derivative of such a function will be homogeneous to the degree k-1.They proved this by simply differentiating both sides of the equation. My problem arises when they differentiate the left hand side (with respect to the first argument as an arbitrary choice). They say the partial differential(of the LHS wrt x1) is:

(∂ƒ(tx1,tx2,…,txn)/∂x1).t

my question is where does the t come from.. ..please bear with me

2. Nov 3, 2014

### Mentallic

They used the chain rule. When you take a derivative of

$$f(g(x))$$ with respect to x, you first take the derivative of f, but then you need to multiply by the derivative of g.

$$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$$

So

$$\frac{d}{dx}f(tx)=f'(tx)\cdot \frac{d}{dx}tx = f'(tx)\cdot t$$

assuming t is independent of x (constant).

3. Nov 3, 2014

### Sidney

thank you :) I don't know why it seems so obvious now ..I did think of the t as being a result of the chain rule but for some reason the way they wrote it down made no sense to me and had me stuck...I think it's because they wrote (∂ƒ(tx1,tx2,…,txn)/∂x1) which to me means with respect to x( i.e. ∂x1) and not with respect to the change in the intermediate function(tx) and so it came across as meaning the complete derivative of ƒ1 encompassing all the intermediate processes..

the way you have your functions written down is so neat. If you don't mind me asking what did you use because the way I'm doing it takes forever, is very messy and I can't write in fraction form

Last edited: Nov 3, 2014
4. Nov 3, 2014

### Mentallic

Things often become clear again when it's explained in simple terms :)