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Let's take a parabola $y=x^2$. Let's take it's derivative for $x=1$: enter image description here

We take the derivative of $y=x^2$, and find that as the x-value on the parabola approaches $x=1$ the derivative is 2. $\lim _{h \to 0} = 2$ at point $x=1$ on the parabola.

My question here is why the once we find the derivative, we draw a line with that slope, over that point. Once we find the derivative of $x=1$, we draw a line with gradient 2 over (1,1).

But this makes no sense to me because of 2 reasons:

  1. Isn't the derivative not the slope at a point, but the slope between 2 points as one point gets arbitrarily close to another? So since it fundamentally requires 2 points to do, how can we just draw a line with that slope over 1 point? If there was a slope at the single point, then it would result in the denominator 0, leading to an undefined result!

  2. How can we just assume that the slope at the point is 2, because the derivative as $x+h$ approaches $x$ is 2? Just because the slope is getting closer and closer to 2 as the $x+h$ gets arbitrarily close to $x$ doesn't mean the slope at $x$ is actually going to be 2!

So with all this, can someone explain to me how we can still draw tangent lines that have the slope given by the derivative at a given point. How can we do so when:

  1. The derivative doesn't give the slope at a point, but the slope between 2 points $x$ and $x+h$ when $h$ is arbitrarily small. If there was a slope at the single point, then it would result in the denominator 0, leading to an undefined result!

  2. How can we just assume that the slope at the point is 2, because the derivative as $x+h$ approaches $x$ is 2? Just because the slope is getting closer and closer to 2 as the $x+h$ gets arbitrarily close to x doesn't mean the slope at $x$ is actually going to be 2!

Thank you so much! By the way, can you try to give the explanation without epsilon delta proofs, and just at the level someone learning Khan Academy Calculus?

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    $\begingroup$ The short answers? That is how derivatives are defined, using limits, and in doing so we avoid the frustrations caused by "division by zero." Why do we do it? Because it is incredibly useful to do so. $\endgroup$ – JMoravitz Sep 4 '18 at 13:58
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    $\begingroup$ "Why Is The Derivative At A Point Drawn As A Tangent Line?" It is the slope of the tangent line at that point. $\endgroup$ – Mauro ALLEGRANZA Sep 4 '18 at 13:59
  • $\begingroup$ Wait but isn't the derivative the slope of (f(x)-f(x+h))/x-(x+h) as h approaches 0? So then how is it the slope the tangent line at that point? it's the line between x and (x+h) as h approaches 0? $\endgroup$ – Ethan Chan Sep 4 '18 at 14:03
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    $\begingroup$ I encourage you to listen to Herb Gross explain this concept in this video at the timestamp provided, and if you have time I encourage you to watch the entire video series. It should be great supplementary material for whatever course you are taking, or even a good replacement for whatever course you are taking. $\endgroup$ – JMoravitz Sep 4 '18 at 14:06
  • $\begingroup$ See also the post : Definition of a Tangent line and Derivatives and the post : Why is derivative is slope of tangent line ? $\endgroup$ – Mauro ALLEGRANZA Sep 4 '18 at 14:07
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In the example above you can negate your worry by looking at it in a simplified way:

$\frac{f(x+h) - f(x)}{x - (x+h)} = \frac{x^2 +2xh +h^2 - x^2}{h} = 2x + h$

The limit of this as $h \to 0 = 2x.$

Hopefully this at least allays your worry of dividing by 0.

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As long as you are dealing with two points the slope is the slope of secant which is not the derivative.

The derivative at a point is found by taking the limit of the slope of secant as the second point approaches the first one so the secant line approaches the tangent line.

Therefore the derivative is the slope of the tangent line and it is a limit.

We are taking the limit to make sense of what seems to be a $0/0$ which does not make sense by itself.

When you are driving at the speed of $75$, that $75$ is the instantaneous velocity,which means the ratio of distance and time when both distance and time approach $0$ at that instance.

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There are other formulas for finding the derivative, for instance $lim$ $h$->$0$ $(f(x+h)-f(x-h))/(2h)$. Under this conceptualization, you choose two points on either side the point you want the 'slope' at, that are the same distance away. The claim that as h approaches $0$ the limit approaches 2 (for y=f(x)=x^2, and x=1) means that as you move the two points on either side closer to it, the slope between them gets closer and closer to the 'limit'. $lim$ $h$->$0$ $(f(x+h)-f(x-h))/(2h)$ = $lim$ $h$->$0$ $((x+h)^2-(x-h)^2)/(2h)$ =

$lim$ $h$->$0$ $(x^2+2xh+h^2-(x^2-2xh+h^2))/(2h)$ = $lim$ $h$->$0$ $(4xh)/(2h)$ =

$lim$ $h$->$0$ $2x = 2x$

The point of saying the limit as h approaches zero of 4xh/h = 2x is because 'if you can't divide by zero' then obviously this equation isn't defined at zero. But if you have a function, and that function 'looks like a point is missing', say at x=4, y looks like it should be 3, then it seems reasonably to extend the function to include that point, or say that the limit as x approaches 4, of y, is 3.

How can we just assume that the slope at the point is 2, because the derivative as x+h approaches x is 2? Just because the slope is getting closer and closer to 2 as the x+h gets arbitrarily close to x doesn't mean the slope at x is actually going to be 2!

To the extent that the limit is about what the slope would be if h was zero, since 4xh/(2h)=2x for all non-zero x, then why shouldn't it equal 2x when h equals zero? If you go the other direction, 2x=4xh/(2h) then you are removing a point instead of adding one. Some limits admittedly aren't about this. The limit as x->infinity (or 'increases without bound') of 1/x = 0 stands in contrast to the fact that there is no value of x for which 1/x=0.

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