# Definition of differential of multivariable function

I have this definition: $f:R^n → R^m$ is differentiable at $a∈R^n$, if there exists a linear transformation $μ:R^n→R^m$ such that

$\lim_{h \to 0} \frac{|f(a+h)-f(a)-\mu(h)|}{|h|} = 0$.

My questions are what's the linear transformation $μ(h)$ for? What does it mean and where does it come from? Why is it necessary?

Can anyone explain the definition to me a bit better? Thanks

• Well, $\mu$ is the actual derivative of $f$, so it's very much necessary when talking about differentiability. – Hans Lundmark Jun 20 '17 at 12:39
• I have seen this expression before. Of what I can tell you is that $\lim_{h\to 0}\frac{\mu(h)}{|h|}$ is the derivative function which implies the expression equals 0. – Pedro Gomes Jun 20 '17 at 12:39
• The idea is that although $f$ might be a complicated nonlinear function, at least $f$ has the nice property that near $a$ the function $f$ can be approximated well by a linear function. And linear functions are easy to work with and to understand. – littleO Jun 20 '17 at 13:27

One can talk about these things in a denominator-free way. The essential point is the following:

Assume that a point of interest $p$ in the domain of $f$ is given, and you want to know how the values of $f$ behave in the immediate neighborhood of $p$. If $f$ is a nice function then moving away by $h$ from $p$ results in an increment $f(p+h)-f(p)$ which is in first approximation a linear function of the displacement vector $h$: $$f(p+h)-f(p)\approx Ah\qquad(h\to0)\ .\tag{1}$$ Such a statement only has real content when the error implied by the $\approx$ sign is for $|h|\ll1$ essentially smaller than the term $Ah$ appearing in the formula $(1)$. Now in most cases $Ah$ will be of order of magnitude $|h|$. We therefore require that $$f(p+h)-f(p)=Ah+o\bigl(|h|\bigr)\qquad(h\to0)\ .\tag{2}$$ If $(2)$ holds in a given situation then $f$ is called differentiable at $p$, and the linear map $A$ (it is uniquely determined) is called the derivative or tangent map of $f$ at $p$.

The linear map $\mu$ is the derivative of $f$ at $a$. It's the best linear approximation of $f$ near $a$.

For a detailed answer look here.

I'll try to give you an equivalent definition, which I think is a bit clearer, I will leave the equivalence to you in first instance. I might add more explanation later.

First let's consider the case that $n=2$ and $m=1$, (actually what I will say works equally well for arbitrary $n$, but $n=2$ is the simplest case that is new to us).

Let me try to be concrete, suppose we have a function $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$, and we want to compute the derivative of $f$, say at $0 \in \mathbb{R}^{2}$. That is, we want to figure out how the output of $f$ changes if we change the input. The idea is to use the fact that we know how to differentiate functions from $\mathbb{R}$ to $\mathbb{R}$. Suppose we have any $v \in \mathbb{R}^{2}$, then we can get a function from $\mathbb{R}$ to $\mathbb{R}$ as follows \begin{align} f_{v}: \mathbb{R} &\rightarrow \mathbb{R}, \\ t &\mapsto f(tv). \end{align} This function is (a reparametrization of) the restriction of $f$ to the line spanned by $v$. We know how to differentiate such a function with respect to $t$ (it might not be differentiable, in which case $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ is not differentiable). We thus get a number $$D_{v} f := \frac{\text{d}}{\text{d}t}\bigg|_{t=0} f(tv),$$ which tells us how quickly the output of the function $f$ changes if we vary the input along the line spanned by $v \in \mathbb{R}^{2}$.

What was described above makes sense for any vector $v \in \mathbb{R}^{2}$, so we have a map from $\mathbb{R}^{2}$ to $\mathbb{R}$, \begin{align} \mu: \mathbb{R}^{2} &\rightarrow \mathbb{R}, \\ v &\mapsto D_{v}f = \frac{\text{d}}{\text{d}t}\bigg|_{t=0} f(tv). \end{align} Now, I claim that the map $\mu$ is linear, and furthermore satisfies the equation that you wrote down (with $a=0$): $$\lim_{h \rightarrow 0} \frac{\|f(h) - f(0) - \mu(h)\|}{\|h\|} = 0.$$ Note that $h \in \mathbb{R}^{2}$. It's up to you to show that $\mu$ satisfies this equation, (and that any $\mu$ that satisfies this equation is the derivative). (I might add some steps to show this later).

Now I have claimed that $\mu: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is the derivative of $f$ at $0$ (or $a$), I should maybe tell you a bit about what it would look like in the case that $m=n=1$. In this case we see (exercise!) that for $s \in \mathbb{R}$ $$\mu(s) = s \frac{\text{d}}{\text{d}t}\bigg|_{t=0} f(t).$$

Like remarked above, I have not really used the fact that $n=2$ and the entire story holds equally well for arbitrary $n$. For higher $m$, we should view a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ as $m$ functions $f_{i}: \mathbb{R}^{n} \rightarrow \mathbb{R}$.