I am trying to understand what exactly a derivative is. I understand the total derivative is a linear map. But I don't understand what happens to the idea of a rate.

In high school calc, one is taught the derivative is a rate of change. For example, $$\frac{dx}{dt}$$

But for a vector function like $$f(x,y) = xy^2 \mathbf{\hat{i}} + x^5 \mathbf{\hat{j}} + \sqrt{y}\mathbf{\hat{k}}$$

What does rate of change mean if I take the total or partial derivative of this?

Should I think of a partial and total derivative like a collection of rates?

  • $\begingroup$ the partial derivative is the rate of change in a specific argument, if you fix the other arguments. For a given direction you can you the total derivative to get the rate of change in that direction. $\endgroup$ – user251257 Jul 19 '15 at 21:04
  • $\begingroup$ I think the best way to think about the derivative is:$ f (x+\Delta x) \approx f (x) + f'(x) \Delta x $. The derivative gives us a local linear approximation to $f $ at $ x $. $\endgroup$ – littleO Jul 19 '15 at 21:29
  • $\begingroup$ While you're pondering this, make absolutely sure you understand the difference between partial and total derivatives. There are situations in which both can be taken but result in different values! (e.g. consider differentiating $y(x, t)$ with respect to $t$ but where $x$ also depends on $t$.) $\endgroup$ – Mehrdad Jul 19 '15 at 23:08

No, you shouldn't think of the total derivative as a rate of change. The generalisation of the concept of the derivative follows from another perspective of understanding it.

One can see it is a rate of change from the following:

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ which is the same as $\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x} $.

However, this leads to some deeply related concept, approximating through a linear function. (You'd see this easily if you knew some Taylor series)


This means that

$f'(x)h=f(x+h)-f(x)+\epsilon(h)$ where $\frac{\epsilon(h)}{h}\to0$ as $h\to0$.

This means that $f'(x)h$ is the best linear approximation of $\Delta y$ in a neighbourhood of x.

Now, you can also see $f'(x)h$ not as a number, but as a linear transformation in $h$. This is because we have linear approximations by a line in $\mathbb{R}^1$ whereas the natural generalisation of linear functions is linear transformations.

Now, through the definition of the total derivative, the total derivative $A$ of $f$ at $x$ is the best approximation that satisfies

$f(x+h)-f(x)=A(h)+\epsilon(h)$ where $\frac{|\epsilon(h)|}{|h|}\to0$ as $h\to0$.

That is $A(h)$ is the best linear approximation of $\Delta y$ in a neighborhood of $x$.

As for the partial derivatives, you can think of them like rates of change because all variables are fixed except one, that is you can think of the rate of change of the function as we move in one direction.

I hope this helps.

  • $\begingroup$ So is it meaningless to speak of "rate of change" of a vector function? Why? $\endgroup$ – Stan Shunpike Jul 20 '15 at 1:05
  • $\begingroup$ It is not meaningless if you're going to take a certain direction. There isn't much of a direction in real-valued functions, but taking a direction works. (Hence the last paragraph) You can also take directional derivatives. However, for the total derivative, explaining it as a rate of change is neither intuitive nor meaningful. $\endgroup$ – Hasan Saad Jul 20 '15 at 2:32

Given a function, $f$, from $R^n$ to $R^m$, we can define the "derivative" of $f$, at a given point, to be the linear function, $y= Ax+ b$, from $R^n$ to $R^m$ that "best approximates" $f$. In a given coordinate system for each of $R^n$ and $R^m$ that can be represented by an $m \times n$ matrix. In particular, if $f$ is a "vector valued function of a single variable, $f(t)= \langle a(t), b(t), c(d)\rangle$, then its derivative is $f'(t)= \langle a'(t), b'(t), c'(t)\rangle$. If $f$ is a "real valued function of several variables, $f(x, y, z)$, its derivative is the gradient, $\nabla f= (\partial f/ \partial x, \partial f/\partial y, \partial f/ \partial z)$.

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    $\begingroup$ I like this answer, but I had to apply MathJax formatting to it (see MathJax basic tutorial and quick reference) in order to read it. I hope the formatting is accurate; you can edit it again if I introduced an error. $\endgroup$ – David K Jul 19 '15 at 22:19

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