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What the meaning of $dx$? Does it refer to something known as a differential (whats that?). Why is it used in integration when we put: $$\int f(x) dx$$

Does it refer to the derivative of something?

What exactly is dx? I don't get it.

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  • $\begingroup$ I tried looking at that but its not what im looking for $\endgroup$ – K Split X Dec 8 '16 at 2:27
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Calculus is all about how to measure the slope of any arbitrary line,
especially curved ones.

Consider $y = 2x^2$.

If you used the "normal" method to get the slope you would pick two points (lets pick $(1,2)$ and $(3,18)$ for this example) and then you would make a ratio of the "rise" (the difference in the $y$ values) and the "run" (the difference in the $x$ values)

If you did this you would have a slope of:

$ m = \frac {y_2-y_1}{x_2-x_1} = \frac {18-2}{3-1} = 16/2 = 8$

The problem with this method is that it produces the wrong answer.
The only time it's right is for a straight line. For example, pick two different points: $(2,8)$ and $(3,18)$

Then you would have this slope:

$m = \frac{y_2-y_1}{x_2-x_1} = \frac {18-8}{3-2} = 10/1 = 10$

A curved line is not straight, so the slope will never be right except for one thing: the closer the two points are to each other, the more accurate the slope is. The curve gets flatter and flatter as the two points get closer and closer. When they get infinitely close to each other we get the most accurate answer because essentially the points are so close to each other that there is no room for any curvy bits.

If we define "dx" to be the difference between two x-values that are infinitely close to each other (an infinitely small difference in x values), and we define "dy" to be the difference between two y-values that are infinitely close to each other (an infinitely small difference in y values), then we can pick two infinitely close points and do this:

$m = \frac{y_2-y_1}{x_2-x_1} = \frac {dy}{dx} $

So $\frac{dy}{dx} $ is the slope of a line. If we use the rules of calculus to "differentiate" our equation (using the mythical d function):

$y = 2x^2$ $d(y) = d(2x^2)$ $dy = 2 d(x^2)$ $dy = 2*2x*d(x)$ $dy = 2*2x*dx$ $dy = 4x*dx$

We find that an infinitely small difference in y can be measured with this equation: $dy = 4x*dx$. But if we rearrange it slightly:

$dy = 4x*dx$ $\frac{dy}{dx} = 4x*dx/dx$ $\frac{dy}{dx} = 4x*1$ $\frac{dy}{dx} = 4x$

We find that the slope of $y = 2x^2 $is $4x$. Notice that the slope is not a number; it actually changes depending on where in the graph we are; you can see that the slope changes by graphing $y = 2x^2.$

So the slope at any point on the graph can be found with this equation because the two points that we use to calculate with are infinitely close together (for all intents and purposes they are the same point)

And since we know the definitions of dx and dy, we could say that the slope at any point equals an infinitely small difference in y (dy) divided by an infinitely small difference in x (dx). This is absolutely true. And for a straight line graph it is the same as taking the difference of any two points.

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  • $\begingroup$ This is a brilliant explanation. Thank you. $\endgroup$ – K Split X Dec 8 '16 at 20:17

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