In this video, 3Blue1Brown shows how one can derive the derivative geometrically. Around 2 min 30 sec, he shows how the derivative of $f(x)=x^2$ is $f´(x)=2x$ by starting with a square.
The key idea is that the change in $f(x)$, denoted $df=dx^2$, captures the change in area. This change has three components: the two rectangles and the tiny rectangle. One obtains the area of one rectangle by multiplying the two sides, that is $xdx$. The area of the tiny area is $dxdx$. Thus, the change is area is then $dx^2=2xdx+dxdx$. As the derivative is defined as the ratio between change in the input (i.e., $x$) and the output (i.e., $f(x)$), we solve for $dx^2/dx$. By algebraic manipulation, that quantity is $dx^2/dx=2x+dx$. When we let $dx$ approach zero (i.e., taking the limit) in this equation we are left with $dx^2/dx=2x$. Thus, we derived $f´(x)=2x$.
In a similar way, we may derive $f(x)=x^{1/2}$. 3Blue1Brown leaves this as a challenge, and the most common way is to now set the area to $x$ and the sides to $x^{1/2}$. So the area changed is now defined as $dx$, and to obtain the geometric pieces we follow the exact same geometric logic: $dx=2x^{1/2}dx^{1/2}+dx^{1/2}dx^{1/2}$. By algebraic manipulation, we obtain $dx^{1/2}/dx=1/2x^{-1/2}$. See the following link for a complete solution (see particularly Armin Meisterhirn's solution for geometric version).
Now to my question. While I understand that by re-writing the problem of finding $f(x)=x^{1/2}$ by setting $x$ to represent the outcome (the total area), if feel unintuitive to stating the problem like this. It forces us to start the problem by seeking to find the area changes via $dx=...$, which we usually define as the input -- now it is defined as the output. It feels more intuitive to starting with the actual output, and defining the total area in terms of $x^{1/2}$ and not $x$. And then focus on $dx^{1/2}=...$, but what would such an area be? Inspired by the fact that $f(x)=x^2=xx$ that lends itself to the two sides of a square, why not write $f(x)=x^{1/2}=x^{1/4+1/4}=x^{1/4}x^{1/4}$. And so we set the sides of the square to $x^{1/4}$. That would enable us to write down the more intuitive expression $dx^{1/2}=2x^{1/4}dx^{1/4}+dx^{1/4}dx^{1/4}$. However, I was unable to manipulate this expression algebraically, to make it equal to $dx^{1/2}/dx=1/2x^{-1/2}$.
The key problem is that I could not obtain $dx$ from $dx^{1/2}=2x^{1/4}dx^{1/4}+dx^{1/4}dx^{1/4}$. I can factor out one, $dx^{1/2}=((2x^{1/4}dx^{1/4})/ dx +(dx^{1/4}dx^{1/4}/dx))dx$, and then $dx^{1/2}/dx=((2x^{1/4}dx^{1/4})/ dx +(dx^{1/4}dx^{1/4}/dx))$ but such a move creates an ugly and complicated expresion for I am not sure what the limit is.
My question is two-folded: (1) do you see a way to continue along my more intuitive path algebraic, and (2) if not, why does my intuitive method not work?
Might it be that exponential does not lend themselves to a generalized geometric solution? 3Blue1Brown hints that this is the case in the follow up video. But then what is the limitation of this approach to the derivative (it can be used for sin, cos, and other complicated functions).
My problem is related to Derivative of Square Root Visual
