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enter image description here

Start by assuming that function curves are made of an infinite amount of lines (i.e. look at the image above but instead of approximating it using a finite number of lines, use infinite lines). This intuition is supported by the arc length formula (integral of infinite "line approximations").

Let's call the point at the top ("the highest point") point $x$. Refer to the $n^{th}$ point before that as $x-n$ and the $n^{th}$ point after that as $x+n$. I want to calculate the derivative at point $x$. In other words I want to calculate the slope of the line connecting point $x$ and $x+1$.

I can start be drawing a secant line from point $x$ to $x+n$ and using the slope to approximate the derivative. As $n$ approaches $1$, you get better approximations of the derivative. As $n=1$ you exactly have the derivative.

Now draw a secant line from point $x$ to $x-n$. As n approaches 1 you get infinitely better approximations of the slope of the line between $x$ and $x-1$, but not $x$ and $x+1$. No matter how close to n=1 you get you will never get the slope of the line that connects $x$ and $x+1$ (which is what the derivative is: the rate of change of the function between a point and the next infinitesimal point).

So why is the derivative the limit:

$$\lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$

instead of:

$$\lim _{h \rightarrow 0^+} \frac{f(x+h) - f(x)}{h}$$

Since using the above intuition shows that the left hand limit doesn't give you the slope of the line between $x$ and the next infinitesimal point.

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    $\begingroup$ We want the geometric intuition of the limit to be true from any side of $\;x\;$ ... $\endgroup$ – DonAntonio Jan 18 '14 at 20:15
  • $\begingroup$ I guess your intuition is missing something: for a function to be differentiable, you want the rate of change from $x-1$ to $x$ to be the same as of $x$ to $x+1$. You don't want the derivative to change dramatically at some point $\endgroup$ – M Turgeon Jan 18 '14 at 20:16
  • $\begingroup$ @DonAntonio Well yeah, but why doesn't this intuition allow that? There has to be some truth to it since its derived from the arc length formula. $\endgroup$ – dfg Jan 18 '14 at 20:18
  • $\begingroup$ Uh, @dfg? The "arc length formula"? I don't see it clearly...anyway, if your intuition serves you well from the right I can't see how won't it serve you well from the left...:) $\endgroup$ – DonAntonio Jan 18 '14 at 20:20
  • $\begingroup$ @MTurgeon But the derivative doesn't change drastically. Any amount of difference between the next slope and the previous slope renders the limit "wrong" according to this intuition. If you assume that the previous slope is always equal to to the next "slope", then all curves would have to be lines. (constant derivative) $\endgroup$ – dfg Jan 18 '14 at 20:20
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It is an interesting question. Your understanding of the situation is quite good, you are just missing a small element.

Suppose your function is $$f(x) = \left| x \right |$$

enter image description here

Using your technique, the limit from the right, you find that the slope at $x=0$ is simply $1$. But what if you use the same technique starting from the left? You get that the slope at $x=0$ is $-1$. This is a contradiction. You find that the slope at a certain point can be two different things, but this is far off the general definition of the tangeant line and its slope!

In fact, in the case of $f(x) = \left| x \right |$, the derivative at $x=0$ is said to be undefined, and you can easily understand why by taking a look at the graph. But, in some cases, you might only be interested in what is going on with the limit from the right, or maybe with the limit from the left.


Added:

A limit $\lim\limits_{x\rightarrow a} f(a)$ is only defined if $\lim\limits_{x\rightarrow a^+} f(a) = L$ and $\lim\limits_{x\rightarrow a^-} f(a) = L$.

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  • $\begingroup$ But isn't the definition of a derivative the rate of change between an infinitesimal point and the next infinitesimal point? Why do you care about the previous rate of change? $\endgroup$ – dfg Jan 18 '14 at 20:25
  • $\begingroup$ Why the right point rather than the left? The general limit is only defined if both left and right are defined and give the same result. If you only want to know about what is going on at the right, you specify it. $\endgroup$ – Olivier Jan 18 '14 at 20:27
  • $\begingroup$ I suppose you could define the derivative as the left point. But why would the "right" point and "left" point change always be the same? If the function always changes at the same rate to the next point as it did from the previous point doesn't that imply the function is a line? $\endgroup$ – dfg Jan 18 '14 at 20:30
  • $\begingroup$ It's a good concern, but in fact that the slope is the same at the left and at the right of a point $x$ doesn't imply that the function is a line. It implies that the function looks very much like a line at the infinitesimal scale at the point $x$, that is that a line would very well approximate the function at a small scale. $\endgroup$ – Olivier Jan 18 '14 at 20:33
  • $\begingroup$ And the derivative is defined as the general limit, that is it has to be defined both with the limit from the left and with the limit from the right. Otherwise, we can have a situation where a "spike" in a function has a slope. $\endgroup$ – Olivier Jan 18 '14 at 20:34
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As others have noted, your definition means that $f(x) = |x|$ is differentiable at zero. But why is the positive direction any more important than the left? You could define a "right derivative" like this if you wanted to, it's just that the two-sided derivative has nicer properties.

You say "Start by assuming that function curves are made of an infinite amount of lines". This is actually really close to the (in my opinion) best way to think about the derivative: as the linear function that best approximates $f$ at that point. You take a small neighborhood around $x_0$, and as that neighborhood gets smaller and smaller, $y - f(x_0) = f'(x_0) (x - x_0)$ looks more and more like $y = f(x)$.

enter image description here

However, for a function like $f(x) = |x|$, there's no line you can draw at $(0,0)$ that approximates it well. No matter how close you zoom in, it will look like this:

enter image description here

Lastly, your intuition about "infinite line segments" is wrong. You're breaking $f$ into finitely many pieces, looking at the slope between $x_0$ and $x_n$, and taking the limit as $n \to 1$. But this depends a lot on how many pieces you break it into, and where you break them! Plus, since it's finite, you could just cut to the chase and find the slope between $x_0$ and $x_1$.

The limit you should be looking at is "the slope between $x_0$ and $x_1$ as the number of slices increases" (well, more accurately, as $|x_0 - x_1| \to 0$). In many cases, this will be the same whether $x_1$ is on the left or the right. But as in the first paragraph, why is right any more interesting than left?

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According to your intuition the function $$ f(x)=\begin{cases}1,&\ x\geq0\\0,&\ x<0\end{cases} $$ is differentiable everywhere, while $$ g(x)=\begin{cases}1,&\ x>0\\0,&\ x\leq0\end{cases} $$ is not.

In particular, in your intuition a function can be differentiable at a point while failing to be continuous.

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  • $\begingroup$ But what exactly about the intuition is wrong? $\endgroup$ – dfg Jan 18 '14 at 20:23
  • $\begingroup$ Why do you care about how the function changed from the previous point to the current point? I thought the derivative is a measure of how the function is going to change "next"? $\endgroup$ – dfg Jan 18 '14 at 20:27
  • $\begingroup$ That you are looking at the slope from just one side. You make that choice when you say " I want to calculate the slope of the line connecting point x and x+1". As you can see from the example, that leads you to a notion where a point of discontinuity can a have a slope. Nothing wrong with that, but it is way more limited than the usual notion. $\endgroup$ – Martin Argerami Jan 18 '14 at 20:28
  • $\begingroup$ Your assertion that the derivative is a measure of how the function is going to change "next" does not agree with basic intuition. If for example you think of the derivative as the velocity, then it makes a lot more sense to think of it as the change from the previous point to the current one; unless you think that the current velocity of a particle should depend on future behaviour! $\endgroup$ – Martin Argerami Jan 18 '14 at 20:31
  • $\begingroup$ But the change from the previous point to the current one gives you the "old" velocity. The change to the next one tells you how the object is moving now. How can my intuition of "line approximations" be adjusted to agree with the general definition? $\endgroup$ – dfg Jan 18 '14 at 20:35
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Your intuition is indeed flawed. Say we can somehow deal with infinitesimal values. Let $\epsilon$ be such an infinitesimal value, greater than $0$, but smaller than any positive real number. Then the derivative of $f$ at $x$ would be $\frac{f(x+\epsilon)-f(x)}{\epsilon}$ (this is probably what you want to convey with your $f(x+1)$ above), and there would be no need for limits at all. Now indeed for a differentiable $f$ the value of $\frac{f(x-\epsilon)-f(x)}{-\epsilon}$ would differ from $\frac{f(x+\epsilon)-f(x)}{\epsilon}$ only by an infinitesimal value, so it would not matter which one you took. The size of that infinitesimal difference would of course determine the second derivative.

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