# Question about the derivative of a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$

I have been reading some lecture notes, which have been somewhat confusing for me.

What the lecture notes state:

Let $$f:\Omega \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^m$$ be a continuously differentiable function, remembering analysis we know that

$$Df: \Omega \rightarrow Lin(\mathbb{R}^n,\mathbb{R}^m)$$ with $$x_{\ast} \in \Omega \mapsto (Df)_{x_{\ast}}$$, where the linear map $$(Df)_{x_{\ast}}$$ evaluated in $$v \in \mathbb{R}^n$$ can be written as:

$$(Df)_{x_{\ast}} \cdot (v):=Jacf(x_{\ast}) \cdot v=$$ $$\begin{pmatrix} \partial_{x_1}f_1(x_{\ast}) & ... & \partial_{x_n}f_1(x_{\ast}) \\ ... & & \\ \partial_{x_1}f_m(x_{\ast}) & ... & \partial_{x_n}f_m(x_{\ast}) \end{pmatrix} \cdot \begin{pmatrix} v_1\\ ...\\ v_n\\ \end{pmatrix}$$

As far as I can remember $$D(f(x_1,...,x_n))=\begin{pmatrix} \partial_{x_1}f_1(x_{\ast}) & ... & \partial_{x_n}f_1(x_{\ast}) \\ ... & & \\ \partial_{x_1}f_m(x_{\ast}) & ... & \partial_{x_n}f_m(x_{\ast}) \end{pmatrix}$$, but if I want to evaluate the matrix $$Df(x_1,...,x_n)$$ in some point $$p=(p_1,...,p_n)$$, I just take the Jacobi matrix and insert $$p_1$$ for $$x_1$$,..., $$p_n$$ for $$x_n$$.

For example: If $$f(x,y)=(\cos(x),\sin(x),x)$$, then the Jacobi Matrix is $$\begin{pmatrix} -\sin(x) & 0\\ 0 & \cos(y) \\ 1 & 0 \end{pmatrix}$$. If I now want to evaluate it at for example in $$(3,2)$$, I would get:

$$\begin{pmatrix} -\sin(3) & 0\\ 0 & \cos(2) \\ 1 & 0 \end{pmatrix}$$

But, the lecture notes would state, to take the matrix $$\begin{pmatrix} -\sin(x) & 0\\ 0 & \cos(y) \\ 1 & 0 \end{pmatrix}$$ and multiply it by $$\begin{pmatrix} 3 \\ 2 \end{pmatrix}$$ , which yields a different result.

My Question is: Have I been understanding it wrong, or did I misunderstand what the mentioned text means?

• The Jacobian is the linear map (matrix) which approximates the differentiable function. When you multiply that matrix by $(3,2)$ you are rather getting the approximated output value from the matrix. Nov 12, 2023 at 4:26
• Take the function $f(x) = x^2$. Then the (Fréchet) derivative at $x$ is the linear map $h \mapsto 2xh$, or $Df(x)h = 2xh$. In one dimension this is not very exciting. Nov 12, 2023 at 6:31

$$Df$$ has two arguments. The first argument is the point in $$\Omega$$, where you evaluate your Jacobian matrix. This matrix then acts (via multiplication) on your vector of increments $$v\in\mathbb{R}^n$$ to produce the linear part of the increment of your function (a vector in $$\mathbb{R}^m$$).

$$Df$$ is linear in its second argument, so it acts via multiplication by a matrix. It is not linear, in general, in the first argument, so you can consider higher order differentials.

• I am sorry, but I am not sure if I understand what you mean. I remember the definition of a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ being differentiable in some point $\xi \in \mathbb{R}^n$. This is the case if there exists an $n \times m$ matrix $A$, such that $\forall \xi+h$ in some neighborhood $U$ of $\xi$, we can write the increment $f(\xi+h)-f(\xi)$, as $f(\xi+h)-f(\xi)=Ah+r(h)$, with $\lim_{h \rightarrow 0} \frac{r(h)}{||h||}$. Nov 12, 2023 at 4:55
• In what you wrote in the first section, this would mean that multiplying the Jacobi Matrix $A$ with the vector $h$ would give me the vector $f(\xi+h)-f(x)$.Did I understand that correct? Nov 12, 2023 at 4:55
• That is correct. Up to $o(h)$. Nov 12, 2023 at 17:33

I think what you're missing here is that the derivative at a point is not a scalar, or a vector, or even (technically speaking) a matrix. It is a linear function. It's a function in the sense that it acts on the space $$\Bbb{R}^n$$ and produces a vector in $$\Bbb{R}^m$$. We are used to representing (naturally) as a matrix, or even as a scalar in the $$\Bbb{R} \to \Bbb{R}$$ case, but it should be thought of as a function.

You can see this in the notation in your notes: the codomain of $$Df$$ is $$\operatorname{Lin}(\Bbb{R}^n, \Bbb{R}^m)$$, the set of linear functions from $$\Bbb{R}^n$$ to $$\Bbb{R}^m$$.

This space of functions can be naturally represented by matrices. For any $$T \in \operatorname{Lin}(\Bbb{R}^n, \Bbb{R}^m)$$, there exists a unique matrix $$M_T \in \Bbb{R}^{n \times m}$$ such that: $$Tv = M_Tv$$ for all $$v \in \Bbb{R}^n$$. So, for this reason, you will often see derivatives represented as matrices.

In your example, $$f(x, y) = (\cos x, \sin y, x)$$, the derivative at a point, say, $$(\pi/2, \pi)$$, will be the linear map: $$\pmatrix{a \\ b} \mapsto \pmatrix{-a \\ -b \\ a} = \pmatrix{-\sin \frac{\pi}{2} & 0 \\0 & \cos \pi \\ 1 & 0}\pmatrix{a \\ b}.$$ This is represented by the above $$3 \times 2$$ matrix.

The particular linear function you get will depend on the point at which you're taking the derivative (remember, $$Df$$ is a function, and depends on the input from its domain $$\Bbb{R}^n$$, or more precisely, the set of points where $$f$$ is differentiable). So, choosing another point will produce a different linear function under $$Df$$, and a different corresponding matrix.

It's worth paying a bit of attention to the $$\Bbb{R} \to \Bbb{R}$$ case. In this case, the linear map is also from $$\Bbb{R}$$ to $$\Bbb{R}$$, leading to a $$1 \times 1$$ matrix. Linear maps from $$\Bbb{R}$$ to $$\Bbb{R}$$ take the form $$x \mapsto mx$$ for some $$m$$. When we start teaching derivatives, we identify the linear map $$x \mapsto mx$$ more simply by the coefficient $$m$$, which is a real number. This is why, when we take the derivative of a single-variable function, e.g. $$f(x) = x^3 + x$$, we produce a scalar-valued derivative $$f'(x) = 3x^2 + 1$$. When I say the derivative is a function, this is not the function I mean. What we are saying, in terms of the more general structure, is that, at a given point $$x_0$$, the derivative $$Df$$, evaluated at $$x_0$$, is the linear map $$x \mapsto (3x_0^2 + 1)x.$$ This is the linear map, and it will change depending on the point $$x_0 \in \Bbb{R}$$ we substitute in.

So, let's look back at $$f(x, y) = (\cos(x), \sin(y), x)$$. What you computed was the matrix representation of the derivative of $$f$$ at the point $$(3, 2)$$. What the question was asking you to do was to compute the matrix representation at an arbitrary point $$(x, y)$$ (giving us a linear map from $$\Bbb{R}^2$$ to $$\Bbb{R}^3$$), then evaluate this linear map at the point $$\pmatrix{3 \\ 2}$$. This corresponds to simply multiplying the vector by its matrix representation.

As you can see, it's a different question to how you interpreted it. You answered a perfectly reasonable question, but it wasn't the question that was asked.

• Thank you very much! Nov 12, 2023 at 6:23