How to differentiate this expression with respect to t

Question

I don't understand how to differentiate $f(tx_1,\cdots, tx_n)$ with respect to t.

Answer 1

I'll do it for $N=3$ expliciting steps. When you write $f(x_1,x_2,x_3)$ think of $x_1,x_2,x_3$ as placeholders for each corresponding coordinate. Saying it is homogeneous of degree $r$ means that when we have $x_1 = t \overline{x}_1, x_2 = t \overline{x}_2, x_3 = t \overline{x}_3$ then it holds $$f(t \overline{x}_1, t \overline{x}_2, t \overline{x}_3) = t^r f(\overline{x}_1, \overline{x}_2, \overline{x}_3).$$ This is equivalent to saying $$f(t \overline{x}_1, t \overline{x}_2, t \overline{x}_3) - t^r f(\overline{x}_1, \overline{x}_2, \overline{x}_3).$$ Now fix everything except $t$. This is a function of one variable $t$, let's name it $g(t)$ as $$g(t) = f(t \overline{x}_1, t \overline{x}_2, t \overline{x}_3) - t^r f(\overline{x}_1, \overline{x}_2, \overline{x}_3).$$ Differentiating it with respect to $t$, we get $$\frac{dg}{dt} = \frac{\partial f}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial f}{\partial x_2} \frac{dx_2}{dt} + \frac{\partial f}{\partial x_3} \frac{dx_3}{dt} - r t^{r-1} f(\overline{x}_1, \overline{x}_2, \overline{x}_3)=0.$$ What happened here is that we applied the chain rule to the portion $f(x_1,x_2,x_3)$ and usual one-variable differentiation to the portion $t^r f(\overline{x}_1, \overline{x}_2, \overline{x}_3)$ because $\overline{x}_1, \overline{x}_2, \overline{x}_3$ are fixed and therefore we interpret it as a constant multiplying $t^r$, that is, think of it as $kt^r$. Now notice that $$\frac{dx_1}{dt} = \overline{x}_1, \frac{dx_2}{dt} = \overline{x}_2, \frac{dx_3}{dt} = \overline{x}_3,$$ hence $$\begin{align}\frac{dg}{dt} & = \frac{\partial f}{\partial x_1} \overline{x}_1 + \frac{\partial f}{\partial x_2} \overline{x}_2 + \frac{\partial f}{\partial x_3} \overline{x}_3 - r t^{r-1} f(\overline{x}_1, \overline{x}_2, \overline{x}_3) \\ & = \sum_{i=1}^3 \frac{\partial f}{\partial x_i} \overline{x}_i - r t^{r-1} f(\overline{x}_1, \overline{x}_2, \overline{x}_3) \\ & = 0.\end{align}$$ This is valid for all $t \in \mathbb{R}$, so we can choose $t=1$ in order to get the desired result. Doing so gives $$\sum_{i=1}^3 \frac{\partial f}{\partial x_i} \overline{x}_i - r f(\overline{x}_1, \overline{x}_2, \overline{x}_3) = 0$$ or $$\sum_{i=1}^3 \frac{\partial f}{\partial x_i} \overline{x}_i = r f(\overline{x}_1, \overline{x}_2, \overline{x}_3).$$