Why is do we ignore $(dt)^2$ when computing derivatives by expanding the function?

Question

Consider this function:

$s = t^2 + 1$

My question is, in general, while differentiating, why do we ignore the term of second order of smallness? I mean shouldn't that affect our observations?

Say I differentiate it:

$ds/dt = 2t$

(By the rules)

Now, If I were to take it in definition sense, i.e. incrementing each, s and t by a little ds and dt as follows..

$(s + ds) = (t + dt)^2 + 1$

$\implies (s + ds) = t^2 + (dt)^2 + 2\cdot t\cdot dt + 1$

$\implies ds = (dt)^2 + 2\cdot t\cdot dt$

$\implies ds - (dt)^2 = 2\cdot t\cdot dt$

$\implies (ds - (dt)^2)/dt = 2t$

$\implies (ds/dt) -dt = 2t $

Here, Equating $2t$ from both equations, I get,

$ ds/dt = (ds/dt) - dt$

Where, either I have to admit $dt = 0$ or I can be intuitive that dt is a small value. Nonetheless, how can $2t$ be the same even when I'm subtracting something $(dt)$ that indeed isn't too small, because if it were, we would ignore it at other places too. How does it work?

Answer 1

It is because the derivative is defined as a limit, as that change goes to $0$. So under the limit, $\delta t\to0$.

Specifically, the definition of the derivative of this function is $$\frac{ds}{dt}=\lim_{\delta t\to0}\frac{s(t+\delta t)-s(t)}{\delta t}$$

So using this, $$\frac{ds}{dt}=\lim_{\delta t\to0}\frac{2t\,\delta t+(\delta t)^2}{\delta t}=\lim_{\delta t\to0}\left(2t+\delta t\right)=2t$$

Answer 2

A soft answer would be that what we require of the derivative is that it measures the linear part of the change in a function $f(x)$ when we change the value of $x$ to $x+h$ (or traditionally to $x+dx$, but that can be confusing - as you have seen).

So in your example, which is the function $f:x\mapsto x^2$, we have $$ f(x+h)-f(x)=(x+h)^2-x^2=(2x)h+h^2 $$ and the part linear in $h$ is $(2x)h$.

So we have that $f(x+h)$ can be approximated in terms of $h$ very roughly as the constant function $h\mapsto f(x)$, better by the constant+linear function $h\mapsto f(x)+f'(x)h$ and so on.

Note that we don't need to put $h^2=0$ or anything, all we are saying is that the linear part is $(2x)h$.

Of course making all this precise is the work of real analysis.

Answer 3

One point of the derivative is to isolate first order behaviour, but that doesn't mean second and higher order behaviour is wholly ignored by differentiation. The second derivative may also exist, for example, and Taylor's Theorem shows how to estimate or reconstruct a (suitable) function from its derivatives.

Suppose we know the position of an object and we want to know its velocity - the definition of velocity makes it a first order quantity. Quite possibly the velocity is changing - this is acceleration, of course. So differentiation as a tool isolates the mathematical concepts which relate to our intuitive understanding.

Answer 4

Let us work with finite deltas instead of differentials.

With $\Delta t=0.0001$, $$\frac{\Delta s}{\Delta t}=\frac{(t+0.0001)^2-t^2}{0.0001}=2t+0.0001.$$

With $\Delta t=0.00000000001$, $$\frac{\Delta s}{\Delta t}=\frac{(t+0.00000000001)^2-t^2}{0.00000000001}=2t+0.00000000001.$$

With $\Delta t=10^{-43}$, $$\frac{\Delta s}{\Delta t}=\frac{(t+10^{-43})^2-t^2}{10^{-43}}=2t+10^{-43}.$$

…

And with $\Delta t=0$ (so to say),

$$\frac{\Delta s}{\Delta t}=\frac{(t+0)^2-t^2}{0}=2t.$$