Euler's theorem for homogenous function proof inquiry

Question

Hi guys I do not understand how this is true:

Note: $\bar x = x_1, x_2, ..., x_N$

I am guessing what happened in the equality above is that some sort of product rule was applied but I do not understand why this would be the case since the only function which has a variable of $x_j$ is $\partial f(\bar x)$.

Can some explain how they got the second term on the RHS of the equality?

Much appreciated!

Answer 1

Hint: Use the product rule.

$$nf(x)=\sum_i\dfrac{\partial f}{\partial x_i}x_i \implies m\dfrac{\partial f}{\partial x_j}=\sum_i\dfrac{\partial^2 f}{\partial x_j x_i}x_i+\sum_i\dfrac{\partial f}{\partial x_i}\dfrac{\partial x_i}{\partial x_j}.$$

Note, that $$\dfrac{\partial x_i}{\partial x_j}=1$$ only if $i=j$ and is zero in all other cases. Sometimes we say this is equal to the Kronecker-Delta $\delta_{ij}$. So the sum on the right is only consisting of one term for which $i=j$.