I'm trying to prove the derivative of $\sqrt{x}$ using geometry.
So far I've created a square with area $x$ and side lengths $\sqrt{x}$.
The derivative of the function is $\frac{d\sqrt{x}}{dx}$ with $dx$ being the increase in area.
I've set up the equations for the change in f,
$df = 2(\sqrt{x})(\text{ }d\sqrt{x})\text{ }+\text{ }d\sqrt{x}\text{ }d\sqrt{x}$
$df = 2(\sqrt{x})(\text{ }d\sqrt{x})\text{ }+\text{ }dx$
I'm supposed to wind up with $\frac{df}{dx}$ = $\frac{1}{2\sqrt{x}}$
What I end up getting is $\frac{df}{dx}$ = $\frac{1}{1-2\sqrt{x}}$
What am I doing wrong?
Thanks
$\endgroup$ 62 Answers
$\begingroup$Your mistake is that you simplified $\left(\mathrm d\sqrt{x}\right)^2$ as $\mathrm dx$. You're squaring the entire differential, not the function within it. Now, all you need to consider is that as $\mathrm d\sqrt{x} \to 0$, $\left(\mathrm d\sqrt{x}\right)^2 \to 0$ much faster. So, you have
$$\mathrm dx = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right)+\left(\mathrm d\sqrt{x}\right)\left(\mathrm d\sqrt{x}\right) = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right)+\left(\mathrm d\sqrt{x}\right)^2$$
This means $\left(\mathrm d\sqrt{x}\right)^2$ is negligible, leaving you with the desired result:
$$\mathrm dx = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right) \iff \frac{\mathrm d\sqrt{x}}{\mathrm dx} = \frac{1}{2\sqrt{x}}$$
$\endgroup$ $\begingroup$While the answer by KM101 is spot on, I think the mistake Andrew is making is much deeper.
Having watched this particular video by 3B1B myself, I can see where Andrew's solution goes wrong:
Consider the initial example in the video - how to find the derivative of $x^2$. The area of the sqaure he has constructed is $x^2$, with the sides as x. Hence, your f is $x^2$. The change in area you find is df or d($x^2$).
In this case however, $\sqrt{x}$ * $\sqrt{x}$ is the f. Or in other words, x is the f. So, change in area is equal to d($\sqrt{x} * \sqrt{x}$) or dx.
And then as KM101 has written,
$df = dx =2(\sqrt{x})(d\sqrt{x}) + (d\sqrt{x})^2$
Ignoring $(d\sqrt{x})^2$ as it evaluates to be too small,
$dx = 2(\sqrt{x})(d\sqrt{x}) \implies \frac{d\sqrt{x}}{dx} = \frac{1}{2\sqrt{x}}$,
which is what you wanted to find in the first place - the derivative of $\sqrt{x}$ with respect to x.
Hope that helps!
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