Chain Rule for Multivariable Calculus

Question

Suppose I have $f:\mathbb{R^2}\rightarrow\mathbb{R}$ such that $f(t,s(t))$.

If I want to observe how sensitive $f$ is with respect to $t$, my understanding is that $\frac{\partial f}{\partial t}$ does not accurately capture the impact of $t$ on $f$, because $t$ also works through $s(t)$ to impact $f$ as well.

My confusion is that is this in a way pre-composed function? Then, if I want to ask the question how sensitive the function is with respect to a tiny change in $t$, the correct derivative I should go for it "total derivative" not "partial", which is,

$\frac{df}{dt}=$$\frac{\partial f}{\partial t}+\frac{\partial f}{\partial s(t)}\frac{\partial s(t)}{\partial t}$?

Can somebody explain clearly how mapping is actually done, and depending what "question" (i.e. sensitivity) I am interested in, looking at "total" versus "partial" matter? Thanks.

Answer 1

The total derivative at a point $p$ is a linear function that encodes the "sensitivity" a function has in each of its arguments, at the point $p$. So if your function is $f\colon\Bbb R^n\to\Bbb R^m$, and the total derivative exists, at each point $p\in\Bbb R^n$ there is a linear map $df_p\colon\Bbb R^n\to\Bbb R^m$ that tells you how $f$ changes depending on which direction you are going. The partial derivative, however, encodes only the change in one particular direction. So if $df_p$ exists then it contains the information about the partial derivatives $\frac{\partial f}{\partial x_1},\dots,\frac{\partial f}{\partial x_n}$.

In your question, you are looking at a real-valued function with a single argument, so if you know the expression of $f$ and $s$ you can simply compute the derivative. However, it might be that the expression of $f$ or $s$ is yet unknown. In this case we do a detour using multivariable calculus and the chain rule:

Fix functions $f\colon \Bbb R^n\to\Bbb R^k$ and $g\colon \Bbb R^m\to\Bbb R^n$. The total derivatives of $f\circ g,f,g$ are linked by the formula $$d(f\circ g)_{p}(x) = (df_{g(p)} \circ dg_p)(x).$$

In your case, we have $f\colon\Bbb R^2\to\Bbb R$ and $g\colon\Bbb R\to\Bbb R^2$ defined by $t\mapsto (t,s(t))$. The total derivative of $g$ at $t\in\Bbb R$ is the linear map given in the canonical basis by the matrix $$\begin{pmatrix} 1 \\ \frac{ds}{dt}(t)\end{pmatrix},$$ which we also write as $dx + \frac{ds}{dt}(t)dy$.

The total derivative of $f$ at $(x,y)$ is given by the matrix $\begin{pmatrix} \frac{\partial f}{\partial x}(x,y) & \frac{\partial f}{\partial y}(x,y)\end{pmatrix}$. Therefore, using the chain rule we know that the total derivative of $h\colon t\mapsto f(t,s(t))$ at a point $t$ is given by $$dh(t) = \frac{\partial f}{\partial x}(t,s(t)) + \frac{ds}{dt}(t)\frac{\partial f}{\partial y}(t,s(t)).$$ This expression is valid whenever $s$ is differentiable at $t$ and $f$ is differentiable at $(t,s(t))$.

Answer 2

if you study $t \rightarrow f(t,s(t))$, then your function is not defined on $\mathbb{R}^2$ but on $\mathbb{R}$, and you can analyse its derivative on $t$, it will work fine.

The formula you wrote is pretty much nonsense. In order to avoid confusion, note it $f(x,y)$ whenever $t$ is not involved.

Answer 3

If $s(t)$ is known, you can express $f(s,t)$ as $f(t)$ and then differentiate.

Alternatively, you can use the formula you stated but be careful to evaluate the first partial holding s constant and the second partial holding t constant.

$ \frac{df}{dt}= \frac{\partial f}{\partial t}\bigg|_s \frac{dt}{dt} +\frac{\partial f}{\partial s}\bigg|_t\frac{ds}{dt}$

This can be useful e.g. when $s(t)$ is not known and $s$ and $\frac{ds}{dt}$ are known only numerically.