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If you look at a function "infinitely close", the difference between two points is a line:

      __
   __/ 
__/ 

Where each "__" is a point, and "/" is the value of a derivative (assume the two "/"s have different slopes). I have this intuition because a function can be approximated by a line at an infinitely close distance (i.e. "Linear Approximations")

If you use the above graph of the function to graph the derivative, the derivative looks like this:

   _
 _|

So at an infinitely close distance the derivative looks like the second graph above.

But now if you look at the derivative at an infinitely close distance, it looks the the first graph. So how can the derivative look two different ways at an infinitely close distance? It looks like the first graph when you look at it directly at an infinitely close distance, but the second graph if you look at its antiderivative at an infinitely close distance and use that to plot it.

I know this obviously isn't rigourous but what part of my intuition is wrong?

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  • $\begingroup$ The derivative of a linear function looks like a horizontal line... I think you should use some real graphs, for example, $y=x^2, y'=2x$, etc. Also, the tangent line to a function has slope based on the derivative of the function, but is not the same as the derivative of the function. $\endgroup$
    – abiessu
    Jan 15, 2014 at 16:28
  • $\begingroup$ Its not a linear function. Assume the two "/"s have different slopes. $\endgroup$
    – dfg
    Jan 15, 2014 at 16:29
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    $\begingroup$ And note that $y=x^2$ is not a linear function. Perhaps you could add some real, graphed examples of what you mean? $\endgroup$
    – abiessu
    Jan 15, 2014 at 16:30
  • $\begingroup$ But I can't add real graphs because there just intuitions. And I know $x^2$ is not linear, I mean't the function I graphed isn't linear . And finally I get that the tangent value isn't the derivative, I never said that. I said the slope is the value of the derivative. $\endgroup$
    – dfg
    Jan 15, 2014 at 16:36
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    $\begingroup$ You're are mistaking on the form of the derivative. It is not defined on at the angle points on your initial graph, so it can't possible be the second graph. Plus, if you want to work with "intuition", you should at least try to work with real examples, not examples made of _ and /. $\endgroup$ Jan 15, 2014 at 17:10

1 Answer 1

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Functions which look like straight lines when you zoom in are called differentiable functions. Not all functions are differentiable. The function $x \mapsto |x|$ is not differentiable at $x=0$. The graph $y=|x|$ looks like a $\vee$ and, no matter how much you zoom in on $x=0$, it will always look like a $\vee$.

If I think I understand your question then you are thinking of a graph as lots of points which are connected to one another by straight line segments. The below diagram shows $y=\sin x$.

enter image description here

Each one of these straight line segments will have its own slope/gradient. Then we can plot the slopes of these straight line segments on another graph. This will give a "step function":

enter image description here

The point is that we need to look what happens as the points get closer and closer together. In the case of the graph $y=\sin x$ you will have more and more dots and more and more line segments.

enter image description here

The slopes/gradients of the line segments will hold their values for less time (because the line segments are shorter) and the difference between the steps will reduce because the line segments change less from one to the next.

enter image description here

In the end, the points will all come together so that the line segments shrink down to points and we will get the graph $y=\sin x$ back:

enter image description here

In terms of the gradient graph, the height between the steps will shrink to zero and the length of each step will shrink to zero so that we again get lots of points making the curve of the derivative (which happends to be $\tfrac{\mathrm{d}y}{\mathrm{d}x}=\cos x$)

enter image description here

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  • $\begingroup$ This is brilliant. Could you add a third, finer discretization that shows how it kind of curves at the ends? $\endgroup$
    – GPerez
    Jan 15, 2014 at 18:22
  • $\begingroup$ The thing I don't understand is why a function can look both like a "step function" or a "line segment function" if you look really close. For example in your example, you let $f(x) = \sin x$ be the function and when you zoomed in infinitely close it looked like a "line segment function" and its derivative $\cos x$ looked like a "step function". Easy enough. $\endgroup$
    – dfg
    Jan 16, 2014 at 2:16
  • $\begingroup$ But if you let $f(x) = \cos x$ be the function and you zoomed in infinitely close, $\cos x$ would look a "line segment function" and $\sin x$ would look a "step function"! So based on how you approach it, a function can look like a "step function" and a "line segment function"! How is this possible? $\endgroup$
    – dfg
    Jan 16, 2014 at 2:17
  • $\begingroup$ And thanks for the diagrams! Their great, I appreciate it. $\endgroup$
    – dfg
    Jan 16, 2014 at 2:17
  • $\begingroup$ @dfg but that is the point here, that a graph of a differentiable function will never look like a set of discrete lines with different slopes; no matter where you zoom in, if the function is differentiable at a given point then all that will be visible when zooming in on that point is a single line with a single slope. $\endgroup$
    – abiessu
    Jan 16, 2014 at 15:11

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