Taking derivative of function $g: \mathbb{R} \to \mathbb{R}$ defined in terms of $f: \mathbb{R}^{n+1} \to \mathbb{R}$.

Question

Suppose we are given $g(r): \mathbb{R} \to \mathbb{R}$ where $g(r) = f(ry, r^2s)$ for $f: \mathbb{R}^{n+1} \to \mathbb{R}$ where $y \in \mathbb{R}^n, s \in \mathbb{R}$. How do we determine $\frac{dg}{dr}$? I have a feeling some sort of Chain Rule might be useful.

In addition, this problem looks very much like the definition of the directional derivative of $\underline{q}$ in the direction of $\underline{v}$ as $p'(0)$ where $p(t) = q(\underline{a} + t\underline{v})$. Is there a similar geometric intuition for what $\frac{dg}{dr}$ is doing?

Answer 1

I'm supposing $y$ and $s$ are fixed. Then, $g$ is a composition $g=f\circ h$ where $f$ is as advertised and

$h:\mathbb R\to \mathbb R^{n+1}$ is defined by $r\mapsto (ry,r^{2}s)$.

Then, the chain rule applies to say that $$\left ( \frac{\mathrm{d} g}{\mathrm{d} r} \right )_{r=r_{0}}=\nabla f(h(r_{0}))\cdot \left ( \frac{\mathrm{d} h}{\mathrm{d} r} \right )_{r=r_{0}}$$ which is, after calculating the derivatives, $$\sum ^{n}_{i=1}y_{i}\frac{\partial f}{\partial x_{i}}(r_{0}y,r_{0}^{2}s)+2r_{0}s\frac{\partial f}{\partial x_{n+1}}(r_{0}y,r_{0}^{2}s)$$

Answer 2

There is a multivariate chain rule that is somewhat easy to remember that solves this quite nicely. Let $g_1,g_2,\dots,g_m:\mathbb{R}^n \to \mathbb{R}$ be continuously differentiable, and let $f:\mathbb{R}^m\to\mathbb{R}$ be differentiable. If $F:\mathbb{R}^n\to\mathbb{R}$ such that $$F(x_1,\dots,x_n)=f(g_1(x_1,\dots,x_n), g_2(x_1,\dots,x_n), \dots , g_m(x_1,\dots,x_n)),$$ then we have $$\frac{\partial F}{\partial x_i} = \frac{\partial f}{\partial g_1}\cdot \frac{\partial g_1}{\partial x_i} + \frac{\partial f}{\partial g_2}\cdot\frac{\partial g_2}{\partial x_i} + \cdots \frac{\partial f}{\partial g_m}\cdot\frac{\partial g_m}{\partial x_i}.$$ (I learned this recently from Spivak's Calculus on Manifolds, but uses a different notation so it would be nice if someone could double check this.)

For this problem, we only need it with respect to one variable, so we use $$ \frac{dF}{dx} = \frac{\partial f}{\partial g_1}\cdot \frac{d g_1}{dx} + \cdots + \frac{\partial f}{\partial g_{n+1}} \cdot \frac{ dg_{n+1}}{dx}. $$ We get $$\frac{dg}{dr} = D_1f(ry, r^2s)\cdot y_1 + D_2f(ry,r^2s)\cdot y_2 +\cdots + D_nf(ry, r^2s)\cdot y_n+ D_{n+1}f(ry, r^2s) \cdot 2rs,$$ where $D_i f$ is the $i$th partial derivative of $f$. This is the notation used in Spivak's book for partial derivative, but I'm not quite sure if this is standard.