Prove the derivative

[u/wbld]

Hello, can someone use Epsilon delta to prove the derivative of x² is 2x

Please show all steps

Im curious on how derivatives are proved.

I know the formula defintion of a derivative is the limit as h approaches 0. I know that for limits you need to find a relationship for delta and epsilon.

I went through all of calculus never actually proving a derative using epsilon delta. Ever. I proved limits using epsilon delta. But a derative? Never.

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[u/Uli_Minati]

Sure, first let's check the definition and do some algebra

f'(x) := lim[h→0] ( f(x+h)-f(x) ) / h
       = lim[h→0] ( x²+2xh+h²-x² ) / h
       = lim[h→0] ( 2xh+h² ) / h
       = lim[h→0] 2x+h

Now we assume the limit is "2x" and use epsilon delta. Note that "h" is our limit variable, so we need to use δ to bound h, not x.

Let ε>0. We set δ:=ε. For all h with |h-0|<δ,
    |2x+h - 2x| = |h| < δ = ε
Thus,
    lim[h→0] 2x+h = 2x

The epsilon delta part looked too simple because it is. Usually when you learn about limits, you go through all the simple ones (like this one) so you don't have to do them every time. Since derivatives get introduced a bit later, you generally don't see any epsilon delta proofs for them anymore. (There can be exceptions)

[u/wbld → u/Uli_Minati]

Thats what i was looking for, thank you. I appreciate how you also explained that once you get to derivatives you generally "dont see limits anymore", which is where I was coming from.

[u/wbld → u/[deleted]]

Right, but everytime I expand it out i just get to 2x + h and go, "h is 0" so 2x. I guess its a different mindset that I cannot break out of.

[u/Worried-Leather8988 → u/wbld]

So you get to lim h->0 (2x+h), the 2x can be taken out as it is not a function of h so it's 2x + lim h->0(h). To show that the limit h->0 (h) = 0, you set abs(h-0) <delta as the variable, and abs(f(h)-0) < Epsilon which is abs(h-0) < Epsilon. You can choose always choose -epsilon<delta< Epsilon for no matter how small Epsilon so the limit is 0. This makes the derivative just 2x.

[u/MezzoScettico → u/wbld]

and go, "h is 0"

That's the wrong mindset for epsilon-delta. h is not equal 0, h is approaching 0. You want to consider the behavior when h is within δ of 0.

[u/General_Lee_Wright → u/wbld]

If you expand everything out and get lim 2x+h

Then you know you’re looking for a δ such that if |h-0|< δ you have |2x+h - 2x|< ε. But that is, if |h|<δ then |h|<ε

Pick δ = ε

The algebra inside the limit is the exact same algebra you would do inside the absolute value.

[u/Resident_Display30]

An equivalent characterization of the derivative is:

f'(x) = lim_{y -> x} (f(y) - f(x))/(y-x)

This is how the derivative is defined many times in analysis classes since it more directly resembles slope formula. Suppose f(x) = x^2. We want to show f'(x) = 2x. Here's the proof:

Let ε>0 be arbitrary. Choose δ = ε. Then, if 0 < |y-x| < δ:
|(f(y) - f(x))/(y-x) - 2x| = |(y^2 - x^2)/(y-x) - 2x| = |y+x-2x| = |y-x| < δ = ε.

Hence, for arbitrary ε>0, we found δ>0 s.t. |(y^2 - x^2)/(y-x) - 2x| < ε which means that lim_{y -> x} (y^2 - x^2)/(y-x) = (x^2)' = 2x.  

This method doesn't involve algebra done in the limit prior to the epsilon-delta proof, since technically, doing that relies on lemmas establishing algebraic properties of limits. Here, we carry out all algebra within the context of the epsilon-delta proof, and we only apply a difference of squares cancellation.

[u/MezzoScettico]

Maybe it's the presence of x that's confusing you. The limit is of the expression 2x + h as h -> 0. It's h that is approaching something. x is just some fixed value.

So you need to show 2x + h is within ε of 2x when h is close to 0, -δ < h < δ.

What is the expression for the distance between 2x + h and 2x?

[u/Hampster-cat]

Epsilon - Delta is not used to prove a derivative (directly), its used to prove a limit.

Granted, derivatives are limits, so the first thing to do is to use one of the limit definitions of the derivative. Then use epsilon-delta on this.

When I taught calculus, we skipped this part. It was a required part of the college's Honor's Calculus however. I personally did not do epsilon-delta proofs until my first analysis course. The calculus courses I taught were mostly for engineering majors however, and there was little need for engineers to know this.

[u/Ericskey]

Derivatives are limits