Why the differential of a function is a linear transformation?

Question

I try to find some applications of the differential, that weird thing that is a linear transformation, what is the reason of linearity? And how this linearity works, I mean, there's something fundamental about that linearity? And where we can see the importance of the differential, like in physics or another field of application.

Answer 1

Yes, there is something absolutely fundamental about linearity. The whole idea of calculus is to approximate a complicated function by a linear one. You have probably heard the slogan that the tangent line to a function at a point is the best linear fit at that point; this is exactly the connection between the derivative and linearity.

What may trouble you is that what we usually think of as the derivative in Calculus I is not linear. For example, if $f : \mathbb R \to \mathbb R$ is given by $f(x) = x^3$ for all $x \in \mathbb R$, then $f' : \mathbb R \to \mathbb R$ is given by $f'(x) = 3x^2$ for all $x \in \mathbb R$, which is certainly not linear.

The point of multi-variable calculus is that we shift our view: instead of thinking about the derivative as a single function, we think of it as a family of functions, one per point. We'll often write something like ${\operatorname d}f_x$ to emphasise that we are thinking of the derivative function at $x$. So, with $f$ as above, we would say that ${\operatorname d}f_7$ is, not the number $f'(7) = 147$, but the linear function ${\operatorname d}f_7 : \mathbb R \to \mathbb R$ given by ${\operatorname d}f_7(\Delta x) = 147\Delta x$ for all $\Delta x \in \mathbb R$. That is, the derivative in the usual Calculus I sense should really just be thought of as the slope of the derivative in the more general sense.

In Calculus III, you learn to speak of the Jacobian matrix of a multi-variable function; for example, for the change-of-coördinate function $g : \mathbb R^2 \to \mathbb R^2$ given by $g(r, \theta) = (r\cos(\theta), r\sin(\theta))$ for all $(r, \theta) \in \mathbb R^2$, we have the Jacobian matrix $\operatorname{Jac} g(r, \theta) = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{pmatrix}$ for all $(r, \theta) \in \mathbb R^2$. This is just another name for the derivative in the sense above, where we think of specifying a linear transformation by its matrix (with respect to the standard basis of $\mathbb R^2$). For example, we would write ${\operatorname d}g_{(2, \pi/6)}$ for the linear transformation with matrix $\begin{pmatrix} \sqrt3/2 & 1/2 \\ -1 & \sqrt3 \end{pmatrix}$ (with respect to the standard basis), which is to say the linear transformation given by ${\operatorname d}g_{(2, \pi/6)}(\Delta r, \Delta\theta) = (\sqrt3\Delta r/2 + \Delta\theta/2, -\Delta r + \sqrt 3\Delta\theta)$ for all $(\Delta r, \Delta\theta) \in \mathbb R^2$.

You may notice that I have regarded my derivatives as functions of "the same" variables as the original function, annotated with $\Delta$s—hopefully this reminds you of the idea of the differential approximation, according to which we write $f(x + \Delta x) \approx f(x) + f'(x)\Delta x$—but this is really a bit deceptive. In the utmost generality, we have a function $h : M \to N$ between smooth manifolds, and get, for each $m \in M$, a differential function ${\operatorname d}h_m : \operatorname T_mM \to \operatorname T_{h(m)}N$ that is a linear map of tangent spaces. It just happens that, when our manifolds are vector spaces $V$, there is a natural identification $V \cong \operatorname T_vV$ at each point.

For a very slightly more complicated example, we have maps $h_1 : \mathbb R \to \operatorname S^1$ and $h_2 : \mathbb R \to \mathbb C$ given by $h_1(t) = e^{i t}$ and $h_2(t) = h_1(t)$ for all $t \in \mathbb R$. These are the same maps except for the codomain, but their derivatives are quite different.

Namely, for all $t \in \mathbb R$, we have canonical identifications $\mathbb R \cong \operatorname T_t\mathbb R$ and $\mathbb C \cong \operatorname T_{h_2(t)}\mathbb C$, with respect to which ${\operatorname d}h_{2, t}$ is multiplication by $i h_2(t)$. On the other hand, we have to choose identifications of the tangent spaces to $\operatorname S^1$ with $\mathbb R$—one very natural choice is to identify, for each $z \in \operatorname S^1$, $\Delta t \in \mathbb R$ with the tangent vector at $t = 0$ to the curve $t \mapsto z e^{i\Delta t\cdot t}$—and then ${\operatorname d}h_{1, t}$ is the identity map for all $t \in \mathbb R$.

Answer 2

The usefulness of differentials comes from the fact that a linear function is very simple to calculate and that, by definition, the differential of a function $f$ defined on a vector space $V$ at a point $x\in V$, among all possible approximations of $f(x+h)$ by a linear function, is the best possible approximation, in the sense that $$f(x+h)=f(x)+\mathrm df_x(h)+o\bigl(\|h\|\bigr).$$

Answer 3

(Initially a comment.)

At a technical level a differential is linear by definition. At a secondary level because linear maps are simple. For example

1. They scale area by a constant factor, called the Jacobian (even in case of $R^n \to R^m$ is $n\leq m$.

2. Ultimately (1) is responsible for the change of variables formula, the area formula (think formula for surface area in calculus), and co-area formula.

3. Linearity of the derivative also gives the notion of tangent space to manifolds. Assuming our manifold is embedded in $R^N$, the tangent space is the image of the derivative of a local parameterization.