Derivation of the multivariate chain rule

I can't believe I couldn't find this information online, but could someone provide me a good proof of the multivariate chain rule ? \begin{align} \frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} + \frac{df}{dy}\frac{dy}{dt} \end{align}

I found multiple derivation of this results online using differentials and mean value theorem, but they don't look like rigorous to me. Somehow dividing the differential by $dt$ doesn't make it rigorous for my point of view...

This question comes in a more general context where I am trying to understand why deriving a composition is effectively a matrix product. So by understanding this formula, I am able to see why building matrix of derivatives is a good tool to compute derivatives by matrix multiplication.

Thanks !

Presumably we are saying that $f$ is a function of $x$ and $y$ (i.e., $f(x, y)$), which are both functions of $t\ \ $ ($x(t)$ and $y(t)$). So what does it mean to write $df/dt$? This is really the derivative of another function $F$ defined by

$$F(t) = f(x(t), y(t)).$$

Define the function $g$ by $g(t) = (x(t), y(t))$ so that $F(t) = f(g(t)) = f \circ g(t)$.

Recall the multivariable chain rule.

Theorem (Multivariable Chain Rule). Suppose $g\colon \mathbf{R}^n \to \mathbf{R}^m$ is differentiable at $a \in \mathbf{R}^n$ and $f\colon \mathbf{R}^m \to \mathbf{R}^p$ is differentiable at $g(a) \in \mathbf{R}^m$. Then $f \circ g\colon \mathbf{R}^n \to \mathbf{R}^p$ is differentiable at $a$, and its derivative at this point is given by $$D_a(f \circ g) = D_{g(a)}(f) \ D_a(g).$$

You can find a proof of this in, e.g., Calculus on Manifolds (Spivak). Back to the problem at hand: how do we use the chain rule to prove that

$$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}?$$

Well, let's try writing this in terms of a "matrix" product,

$$\frac{df}{dt} = \begin{bmatrix}\dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y}\end{bmatrix}\begin{pmatrix}dx/dt\\dy/dt\end{pmatrix}.$$

But this is exactly what the chain rule states when applied to the function $F = f \circ g$. We have that

• $D_a(f \circ g) = D_a(F) = \dfrac{dF}{dt}$ (evaluated at some point $a$)
• $D_{g(a)}(f) = \begin{bmatrix}\dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y}\end{bmatrix}$ (each term evaluated at $g(a)$)
• $D_a(g) = \displaystyle \begin{pmatrix}dx/dt\\dy/dt\end{pmatrix}$ (each term evaluated at $a$)

where we have assumed differentiability of the maps.