Please explain what a derivative is... (1 Viewer)

[Lina3]

I've been doing the 2U maths and there is just one thing that's driving me crazy...I just cannot understand was a derivative actually is. The books and sites provide very complex definitions and I just want to understand it clearly. What is a derivative?

Solving them is ok but I feel so stupid doing it and not knowing the point.
For example something easy like f(x)=x (squared) is f'(x)=2x right? Then what does that 2x actually mean? Is it an equation of the tangent line or what???

Sorry if I sound stupid, but it's all a mess in my head right now, so could someone please explain it to me???

 

[slyhunter]

The derivative is the gradient of the tangent for any point.

For example, if f'(x)=2x, say at point (2,4), the gradient of the tangent there would be 4.

 

[Lina3]

But then what of this definition?


I know no-one solves derivatives this way, but that's how I was introduced to it...so wouldn't it be saying more than just the gradient of the tangent line at a given point? If there isn't any more to it, well then the lectures I have been watching confused me way too much for some reason >< and I am a complete idiot ...

 

[such_such]

There's basically two ways of finding the derivative. The one above, where there's limits and all that is called differentiating by first principles, and basically that's the one everyone has to know how to do and prove first. It's very long compared to the second one, where I'm guessing you already know what to do

 

[RivalryofTroll]

If you do the questions first then try to actually understand the content or why you're doing it at a diffferent time, I think it might help!
If you know how to do it, you'll understand better on why you're doing it in my opinion.

 

[nightweaver066]

But then what of this definition?
View attachment 24036

I know no-one solves derivatives this way, but that's how I was introduced to it...so wouldn't it be saying more than just the gradient of the tangent line at a given point? If there isn't any more to it, well then the lectures I have been watching confused me way too much for some reason >< and I am a complete idiot ...

You're meant to be introduced to it as it is the first principle of derivatives. From this formula, the derivatives of trigonometric functions, exponentials and logarithms have been derived. It is also from this that the so called power rule (drop the power and minus one) came from.

The derivation for first principles isn't so difficult and if someone does post it here, i'm sure it would make much more sense to you why the definition, that it is simply the gradient of a tangent at any point of a curve, is correct.

Once we differentiate something, we have the first derivative.

Now the first derivative is also called the gradient function because for any x value you put in, you find the gradient of the tangent to that curve at that particular point.

So for the gradient function, f'(x), if we wanted to find the gradient of the tangent at the point where x = 4, you sub x in, i.e. f'(4).

 

[RealiseNothing]

The derivative as a limit

Lets say we have the graph of a function f(x) = x^2

Since this is continually increasing in a non-linear way, we can't find the gradient of the line as it is constantly changing.

However, lets find the gradient between two points of the curve at say (x,f(x)) and (x+h,f(x+h))

Using rise over run we get the gradient of the line connecting these two points to be [f(x+h) - f(x)] / [h] (the run is just h since it's x+h and x)

Now let's move the point on the curve (x+h,f(x+h)) closer to the point (x,f(x)), so that the line connecting to two points gets smaller and smaller. Doing this will make the gradient more accurate as we are approaching the true rise over run between two points.

We let x+h get as close as we like to x so that the gradient of the line connecting the two points is the same as the gradient at the point x. That is, we take the limit of our original gradient as h approaches 0.

The general case is:

f'(x) = limit[h->0] [f(x+h)-f(x)]/[h]

Lets use f(x)=x^2 as an example:

Since f(x) = y it is obvious that f(x+h) = (x+h)^2

Sub into the general formula:

limit h->0 [(x+h)^2 - x^2] / [h]

Expanding the top gives:

[x^2 + 2xh + h^2 - x^2] / [h]

The x^2's cancel out:

[2xh + h^2] / [h]

Cancel h from top and bottom:

2x + h

Now take the limit as h approaches 0:

2x + 0 = 2x

So the derivative of the function f(x) = x^2 is 2x

 

[RealiseNothing]

^^^That is essentially what I said.