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I am just starting Calculus, and am very confused about the idea of derivatives. I get why we need derivatives, because we need to find the slope of a single point in a line, and our conventional method of finding a slope (y1-y2/x1-x2) won't work if there's only one point.

But there are many things about this that confuses me:

  1. How can there be a slope if there's only one point? The whole premise makes no sense to me, finding the slope of a single point. To have a slope, you need a line, but how can you have a line with only one point? So how is it possible to find the slope of one point; and what would that result even mean? What does the slope of a point mean, when only lines can have slopes?

  2. Are the values found by derivatives just approximations? Because from what I read, the solution to the problem above is it find a point that is infinitesimally close, then find the slope of that? So isn't that an approximation?

I understand the "solution to this problem when calculating is to shift delta x to 0 as seen here: enter image description here. But then if we shift delta x to 0, then the base of the fraction would be 0! Then how would it be possible to calculate?

  1. Does that also mean, we can never find the actual slope of a point because all the results we find are just slopes of lines that are very small, but not slopes of actual points?

Can you try to explain this not too rigorously, at the level of someone just starting Calculus? Thanks.

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    $\begingroup$ Given a curve and a point of it, consider a tangent line ... Also, before the notion of derivative, you should be acquainted with the notion of limit. $\endgroup$ – Hagen von Eitzen Aug 8 '18 at 3:35
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    $\begingroup$ The derivative of a curve at some point, is the slope of the curve "at" the point. Not the slope "of" the point. $\endgroup$ – Elements in Space Aug 8 '18 at 3:38
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    $\begingroup$ @ElementsinSpace Okay that makes more sense. But then if it is the slope of a curve at a specific point, where do we start measuring the slope? What point do we choose to start measuring the slope from? And also, if it just the slope of a curve at a specific point, why do we need to do all this complex stuff with finding a point that's infinitely close to it? $\endgroup$ – Ethan Chan Aug 8 '18 at 3:43
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    $\begingroup$ Derivatives are defined using limits. Before anything else, you should familiarize yourself with the rigorous definition of what a limit is. $\endgroup$ – JMoravitz Aug 8 '18 at 3:52
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    $\begingroup$ I recommend watching this video series from the legendary Herb Gross. The video is old and in black-and-white, but the explanations are still as wonderful as ever. His explanations may help provide you with some better intuition and should answer your specific questions. $\endgroup$ – JMoravitz Aug 8 '18 at 3:56
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Not an expert on calculus, but I know enough to hopefully be able to help with your questions.

  1. You're not necessarily looking for the slope of a point, you're looking for the slope of a curve at that particular point.

For instance, in a straight line of the format $f(x)=mx+b$, like you encountered in algebra, the slope is always the same. However, in an equation that curves, like $f(x)=x^2$, the slope or steepness of the line is constantly changing. Calculus allows you to find how steep that curve is at an exact point.

  1. The beauty of calculus is that, because you're dealing with infinitely "zooming in" on a line, it allows you to actually arrive at an exact answer, not just an approximation. If you were just looking at a point very close to that original point, you'd have an approximation, but because it's infinitely close, you can treat it as the same point for the purposes of calculus.

  2. Again, this allows you to find an exact slope of a line at any point on that line.

Don't feel bad -- it is a lot to take in at first and seems contradictory, but give it time, and you'll come to see the beauty and genius of the calculus!

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    $\begingroup$ In attempt to answer your question in the comments, you can choose to measure the slope at any point in the line. For instance, suppose you graph the speed of a car as it accelerates from 0 to 60 mph in 12 seconds. For each second from 0-12 on the x axis, you mark the speed it was going at that second on the y axis. Connect all those points and you'll likely get a curve. You can look at the graph and see what the speed was at any given second; calculus allows you to find the acceleration, or how fast the speed was changing, at any given second. $\endgroup$ – Kevin H Aug 8 '18 at 3:52
  • $\begingroup$ Also, as mentioned above, the concept of limits will probably help in understanding how one can arrive at an exact answer using derivatives. $\endgroup$ – Kevin H Aug 8 '18 at 3:56
  • $\begingroup$ Also, dealing with the denominator being 0 in your edit, let's say we have $f(x)=x^2$. Then $f'(x)=\frac{f(x+dx)-f(x)}{dx}\ = \frac{(x+dx)^2-x^2}{dx}\ = \frac{[x^2 + 2xdx + (dx)^2]-x^2}{dx}$. At this point the $x^2$s cancel out, leaving $\frac{2xdx+(dx)^2}{dx}$. Now you can cancel the $dx$ on the bottom with the ones on top, leaving $2x+dx$. Now there's no denominator. When you shift $dx$ to 0, you're left with the derivative, $2x$. $\endgroup$ – Kevin H Aug 8 '18 at 4:20

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