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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.

Martin Sleziak
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user10478
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  • Could you explain the question instead of linking the video? – Taroccoesbrocco Jun 19 '18 at 13:36
  • I stated the challenge in text, but I think the video is still important as the question is inherently visual in nature. – user10478 Jun 19 '18 at 13:42
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    "but how this equals..." What is "this". I assume this is "the area added to the square". In other words $dx$. So you have $dx = 2\sqrt x(d\sqrt x + (d\sqrt x)^2$. And you are asked what is $\frac {d\sqrt{x}}{dx}$. To put it another way, $2\sqrt x(d\sqrt x + (d\sqrt x)^2$ is the total change in the area. That's not what the derivative is. The derivative is the ratio of the total change in side as compared to the total change in area. – fleablood Jun 19 '18 at 17:31

1 Answers1

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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 $\frac {\Delta \sqrt x}{\Delta x}$.

fleablood
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