Derivative of Square Root Visual

Question

Derivative formulas through geometry | Essence of calculus, chapter 3 (3Blue1Brown): https://www.youtube.com/watch?v=S0_qX4VJhMQ

There is a challenge at 12:23 asking the viewer to arrive at the formula for $\frac{d}{dx}\sqrt{x}$ by considering small changes in the length and area of a square. What is the correct way to approach this?

I tried to copy the technique used to find $\frac{d}{dx}(x^2)$ at 2:25 by letting $u = \sqrt{x}$. This makes the area added to the square $2u \, du + du^2$, which comes out to $$2\sqrt{x}(d\sqrt{x}) + (d\sqrt{x})^2.$$ The second term is negligible, but how this equals $\frac12x^{-1/2}$ is still unclear.

Answer 1

This makes (the area added to the square) $2udu+du^2$, which comes out to $2\sqrt(x)(d\sqrt(x))+(d\sqrt(x))^2$. The second term is negligible, but how (this) equals $\frac 12x^{-\frac 12}$ is still unclear.

"this" = "the area added to the square" $= dx = 2udu + du^2= 2\sqrt(x)(d\sqrt(x))+(d\sqrt(x))^2\ne$ "the derivative of the square root".

"the derivative of the square root" = $\frac {\text{"the length added to the side"}}{\text{"the area added to the square"}}=\frac {\text{"that"}}{\text{"this"}}=\frac {d\sqrt(x)}{dx}$.

===== full answer ====

If I'm giving a challenge to derive $\frac {d\sqrt{x}}{dx}$ by considering small changes in the length and area of the square, I'd .... do just that.

$s = $ side of square .

$A = s^2 = $ area square.

Small change in length $= \Delta s$

Small change in area $= \Delta A$.

So New Area $= A + \Delta A = (s + \Delta s)^2=\text{New Side}^2$

So $A + \Delta A = s^2 + 2s\Delta s + (\Delta s)^2$

Now the thing to keep in mind is that we are thinking of $s$ as
function of $A$ (not the other way around). $s = \sqrt A$. $A$ is
the input and $s = \sqrt A$ is the output [$*$].

Now we want to solve for $\frac {ds}{dA} = \lim\limits{\Delta A\to 0}\frac {\Delta s}{\Delta A}$ expressed in terms of $A$.

So ... we do it.

$A + \Delta A = s^2 + 2s\Delta s + (\Delta s)^2$

$A + \Delta A = A + 2\sqrt{A}\Delta s + (\Delta s)^2$

$\Delta A= 2\sqrt A \Delta s + (\Delta s)^2$
$\Delta A = (2\sqrt A + \Delta s)(\Delta s)$



$\frac {\Delta s}{\Delta A} = \frac 1{2\sqrt A + \Delta s}$.

Take the limit as ${\Delta A\to 0}$ $\frac {ds}{dA} =\lim\limits_{\Delta A\to 0}\frac 1{2\sqrt A + \Delta s}=\frac {1}{2\sqrt A}$.

$\ $

If that was hard to follow or we lost track of the goal we could go
back to the ($*$) part and put it in terms of $x$.


Let $s = \sqrt x = $ side of square.

Let $A = x = $ area of square.
$\Delta s = \Delta \sqrt x$ $\Delta A = \Delta x$. And $(x + \Delta x) = (\sqrt x + \Delta \sqrt x)^2$ Solve for $\frac {\Delta \sqrt x}{\Delta x}$.