# The Unapologetic Mathematician

## The Geometric Meaning of the Derivative

Now we know what the derivative of a function is, and we have some tools to help us calculate them. But what does the derivative mean. Here’s a picture:

In green I’ve drawn a function $f(x)$ defined on (at least) the interval $(0,4)$ of real numbers between ${0}$ and $4$. The specifics of the function don’t matter. In fact having a formula around to fall back on would be detrimental to understanding what’s going on.

In red I’ve drawn the line with equation $y=g(x)=f(2)+f'(2)(x-2)$. This describes a function with two very important properties. First, when $x=2$ we get $g(2)=f(2)$, so the two functions take the same value there. Second, the derivative $g'(x)=f'(2)$ everywhere, and in particular $g'(2)=f'(2)$. That is, not only do both graphs pass through the same point above $x=2$, they’re pointing in the same direction. As they pass through the point, the line “touches” the graph of $f$, and we call it the “tangent” line after the Latin tangere: “to touch”.

So the derivative $f'(x_0)$ seems to describe the direction of the tangent line to the graph of $f$ at the point $x_0$. Indeed, if we change our input by adding $\Delta x$, the tangent line predicts a change in output of $f'(x_0)\Delta x$. Remember, it’s this simple relation between changes in input and changes in output that makes lines lines. But the graph of the function is not its tangent line, and the function $f$ is not the same as the function $g$ defined by $g(x)=f(x_0)+f'(x_0)(x-x_0)$. How do they differ?

Well, we can subtract them. At $x_0$, we get a difference of ${0}$ because of how we define the function $g$, so let’s push away to the point $x_0+\Delta x$. There we find a difference of $f(x_0+\Delta x)-f(x_0)-f'(x_0)\Delta x$. But we saw this already in the lead-up to the chain rule! This is the function $\epsilon(\Delta x)\Delta x$, where $\lim\limits_{\Delta x\rightarrow0}\epsilon(\Delta x)=0$. That is, not only does the difference go to zero — the line and the graph pass through the same point — but it goes fast enough that the difference divided by $\Delta x$ still goes to zero — the line and the graph point in the same direction.

Let’s try to understand why the tangent line works like this. It’s pretty difficult to draw a tangent line, except in some simple geometric circumstances. So how can we get ahold of it? Well instead of trying to draw a line that touches the graph at that point, let’s imagine drawing one that cuts through at $x=x_0$, and also at the nearby point $x=x_0+\Delta x$. We’ll call it the “secant” line after the Latin secare: “to cut”. Now along this line we changed our input by $\Delta x$ and changed our output by $f(x_0+\Delta x)-f(x_0)$. That is, the relationship between inputs and outputs along this secant line is just the difference quotient $\frac{\Delta f}{\Delta x}$!

We know that the derivative $f'(x_0)$ is the limit of the difference quotient as $\Delta x$ goes to ${0}$. In the same way, the tangent line is the limit of the secant lines as we pick our second point closer and closer to $x_0$ — as long as our function is well-behaved. It might happen that the secants don’t approach any one tangent line, in which case our function is not differentiable at that point. In fact, that’s exactly what it means for a function to fail to be differentiable.

So in terms of the graph of a function, the derivative of a function at a point describes the tangent line to the graph of the function through that point. In particular, it gives us the “slope” — the constant relationship between inputs and outputs along the line.

December 28, 2007 - Posted by | Analysis, Calculus