Undergrad Differentiation is Exact or Approximation

[Devil Moo]

Is Differentiation exact or just an approximation?

I am wonder whether this question is meaningful or not. Slope is expressed as "it is approaching to a value as x is approaching 0" so it is inappropriate to ask such question. But when I deal with uniform circular motion, it is very confusing.

Suppose $A$ is constant for vector $\mathbf A$. And the angle between $\mathbf A(t+\Delta t)$ and $\mathbf A(t)$ is $\Delta \theta$.
$\begin{align} \Delta \mathbf A & = \mathbf A (t + \Delta t) - \mathbf A(t) \nonumber \\ | \Delta \mathbf A | & = 2A \sin (\Delta \theta / 2) \nonumber \end{align}$

if $\Delta \theta \ll 1$, $\sin (\Delta \theta / 2) \approx \Delta \theta / 2$
$\begin{align} | \Delta \mathbf A | & \approx 2A (\Delta \theta / 2 \nonumber \\ & =A \Delta \theta \nonumber \\ | \Delta \mathbf A / \Delta t | & \approx A (\Delta \theta / \Delta t) \nonumber \end{align}$

if $\Delta t \rightarrow 0$,
$| d \mathbf A / dt | = A (d \theta / dt)$

But isn't it $| d \mathbf A / dt | = 2A (d \sin (\Delta \theta / 2) / dt)$?

Is $v = r \omega$ not accurate compared with $v = 2r (d \sin (\Delta \theta / 2) / dt)$?

 

[BvU]

Differentiation is exact: it is expressed as a limit and that makes for an outcome without uncertainty.

$| d \mathbf A / dt | = 2A (d \sin (\Delta \theta / 2) / dt)$
On the left you have a differential quotient, but on the right you have a differential of a difference.
With $\theta/2 = \omega t/2$ you do get the same differential $\omega r$.

 

[Devil Moo]

By chain rule,

$\begin{align} \frac {d\sin(\theta/2)} {d(\theta /2)}\frac {d(\theta / 2)} {d\theta} & = \frac {1} {2} \cos\frac {\theta} {2} \frac {d\theta} {dt} \nonumber \\ |\frac {d\mathbf A} {dt} | & = A\cos\frac {\theta} {2} \frac {d\theta} {dt} \nonumber \end{align}$

It seems they are not the same differential.

Also,

$| \frac {\Delta A} {\Delta t} | \approx A\frac {\Delta \theta} {\Delta t}$

when $t \rightarrow 0$, why it will become equality?