Derivation of the multivariate chain rule

Question

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 !

Answer 1

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.

Answer 2

I also found most of the answers in the net circular and assuming the multiplication of two vectors for multivariable derivatives without proving it explicitly.

So I went back to basics and used the definition of derivative which is $\frac{f(x+e) - f(e)}{e}$ when $e \rightarrow 0$, which means $f(x+e) \simeq f(x) + e\cdot \frac{df}{dx}$

When applied to $f(x, y)$, then $f(x+e_{1}, y+e_{2}) \simeq f(x, y+e_{2}) + e_{1} \cdot \frac{df}{dx}\\ \simeq f(x, y) + e_{1} \cdot \frac{df}{dx} + e_{2} \cdot \frac{df}{dy}.(1)$

Also $x = x(t)$ and $y = y(t)$. So, $x(t+e) \simeq x(t) + e \cdot \frac{dx}{dt}\\ y(t+e) \simeq y(t) + e \cdot \frac{dy}{dt} (2)$

(by the way, here you can define a vector $z =[x, y]$ and assume by convention if $z$, $x$ and $y$ are functions of $t$ that $z(t+e) = [x(t+e), y(t+e)] \simeq z(t) + e \cdot [dx/dt, dy/dt]$. Here the array things start coming).

Finally, $f(t+e) = f(x(t+e), y(t+e))$, and as per (2)

$f(t+e) \simeq f(x(t) + e\frac{dx}{dt}, y(t) + e\frac{dy}{dt})$ and as per (1) $f(t+e) \simeq f(x(t), y(t)) + e\cdot\frac{dx}{dt}\frac{df}{dx} + e\frac{dy}{dt}\cdot\frac{df}{dx} = f(x, y) + e(\frac{dx}{dt}\cdot\frac{df}{dx} + \frac{dy}{dt}\cdot\frac{df}{dy}),$

But also $f(t+e) \simeq f(t) + e\cdot\frac{df}{dt}$. So when $e \rightarrow 0$, we got $\frac{df}{dt} = \frac{df}{dx}\cdot\frac{dx}{dt} + \frac{df}{dy}\cdot\frac{dy}{dt}.$ Not very rigorous proof though.