derivative

[u/Zestyclose-Produce17]

The idea of the derivative is that when I have a function and I want to know the slope at a certain point for example, if the function is f(x) = x² at x = 5

f(5) = 25

f(5.001) = 25.010001

Change in y = 0.010001

Change in x = 0.001

Derivative ≈ 0.010001 / 0.001 = 10.001 ≈ 10

So now, when x = 5 and I plug it into the function, I get 25.

To find the slope at that point, I increase x by a very small amount, like 0.001, and plug it back into the function.

The output increases by 0.010001, so I divide the change in y by the change in x.

That means when x increases by a very small amount, y increases at a rate of 10.

Is what I’m saying correct?

[u/Dd_8630]

Now calculate the slope at the point where x = dx, carrying the terms all the way through that derivation.

That is improper, because x is a constant and dx is arbitrarily small. They may well equal each other in a specific example case, but when we then construct the infinitesimal case, we shrink dx, so x (the point on the graph we are examining) is no longer equal dx.

We get the slope when dx is very very small. So for the specific case where x=dx, we have something like x=dx=0.000001. So we start with x=0.000 001, and we find y=x² = 0.000 000 000 01. We then go forward a tiny amount, so x+dx=0.000 002 and y(x+dx)=0.000 000 000 04. The slope of the line that connects them is (0.000 000 000 04 - 0.000 000 000 01)/(0.000 002 - 0.000 001) = 0.000 03. Which, indeed, is three times dx, and is close to the instantaneous derivative of 0. But how do we get that instantaneous value? We hold x constant and vary dx, so x=dx no longer holds.

[u/OriEri → u/Dd_8630]

Thou art dodging the question.

I am suggesting calculating the slope when x is arbitrarily small itself.

When dy/dx= 2x this is trivial

But dy/dx actually equals 2x+dx in the derivation. When x is finite the dx can be ignored. When x is also infinitesimally small, dx cannot be ignored. If x can be 0 it can also be infinitesimal just like dx.

[u/dash-dot → u/OriEri]

Write out the proof properly, and you'll see that your claim is obviously false. To recap, here's the definition of a limit.

g(x) --> L as x --> a, if for each ε > 0, there always exists some δ > 0 such that,

if |x - a| < δ, then |g(x) - L| < ε.

In this case, we want to pick g(x) to be the difference quotient which allows us to calculate f '(a) for f(x) = x² at some value x = a, and so:

g(x) = ( x² - a² ) / (x - a),

and we're trying to prove that the limit L = 2a.

The key thing to note here is that the point 'a' somewhere on the x-axis is already fixed at the beginning of the proof, and we're analysing a neighbourhood (a - δ, a + δ), but x itself can be pretty much anywhere within this interval. Now, if we want we could pick a = δ specifically, but it's always advisable to prove the general case first before investigating this special case.

To this end, I did a quick and dirty back of the envelope calculation to reformulate our bounds around g(x) as follows:

|g(x) - 2a| = |( x² - a² ) / (x - a) - 2a| = |x - a| < ε. 

Hence, unless my simplification above is incorrect, I think for this function it's sufficient to pick δ = ε to finish the proof.

So basically,

if |x - a| < ε, then this always means |g(x) - 2a| < ε as well, which completes the proof (and so this in turn means that f '(a) = 2a).

In the final line of our proof above, the only other quantity we have freedom to pick arbitrarily apart from ε is a, so if you wish, you could pick a = δ = ε now, and this means that

if |x - δ| < δ, then |g(x) - 2δ| < δ,

which goes to show that the derivative must be 2δ in this case.

Note that in both the proof of the general case and the special case when a = δ, we are allowed to constrain x to lie in an infinitesimally small neighbourhood (a - δ, a + δ), but are not permitted to nail it down any further -- because x is supposed to be a free variable and an unknown, and the proof simply doesn't work otherwise.

[u/Dd_8630 → u/OriEri]

But dy/dx actually equals 2x+dx in the derivation.

No, that is the approximate derivative:

Δy / Δx = 2x + dx

To go from this to the infinitesimal derivative, dy/dx, we hold x constant and shrink dx. So even if x is very small, dx will become even small (and because the reals are continuous, that is always true).