If f satisfies f(tx) = (t^n)f(x) for all t and some fixed positive integer n, show that:
gradient f(x) * x = nf(x)
*Note: x represents a vector
Let $\displaystyle \mathbf{x} = (x_1,\ldots,x_k)$ (assuming that the vector is in a k-dimensional space). You are told that $\displaystyle f(tx_1,\ldots,tx_k) = t^nf(x_1,\ldots,x_k)$. Differentiate that equation with respect to t (using the chain rule to differentiate the left-hand side), then put t=1.