Derivative of $\sqrt{x}$ using geometry I am having trouble with a problem given in a video by 3Blue1Brown, to which I have already found a response, here.
My is issue is understanding why the equation $$\mathrm dx = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right)+\left(\mathrm d\sqrt{x}\right)\left(\mathrm d\sqrt{x}\right) = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right)+\left(\mathrm d\sqrt{x}\right)^2$$ was assembled (in the above thread).
In the video, he does say dx = "New area" and by calculating "New area" I did reach to the same equation. My lack of understanding is why dx is regarded as the new area.
Link to the problem in the video.
Thanks :)
 A: Change the area of a square from $A$ to $A+dA$. At the same time, the side varies from $S$ to $S+\Delta S$ and
$$(A+\Delta A)-A=(S+\Delta S)^2-S^2=2S\,\Delta S+\Delta S^2.$$
Hence
$$\frac{\Delta S}{\Delta A}=\frac1{2S+\Delta S}\approx\frac1{2S}=\frac1{2\sqrt A}$$ as the term $\Delta S$ is made negligible.
A: 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($x^2$). 
3B1B wants you take a slightly different approach here and create a square with sides as $\sqrt{x}$.

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 change in area is, 
$df = dx =2(\sqrt{x})(d\sqrt{x}) + (d\sqrt{x})^2$
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!
