Differential calculus - Derivative of √x - Geometric intuition Why we divide the small difference 'd√x' by the difference in area 'dx' Where  we normally divide the difference in area by the small difference ?
My huge confusion is, when we try to find the derivative of x^2 we divide the difference in area 'dx^2' by the small difference dx -> dx^2 / dx . But while i try to find the derivative of √x, i get a correct answer only when i divide the small difference by the difference in area -> d√x / dx ??
Thanks in advance! :)
 A: You are trying to find the rate of change in the side of a square relative to the change in area.  So you have a division, similar to finding other instantaneous rates of change.     
If $x$ is the area and $s=\sqrt{x}$ the side then you are looking at what happens when you increase the area slightly.  Writing $d\sqrt{x}$ at this stage may be more confusing than something like $\delta s$
You get the change in area to be $\delta x = 2 s \, \delta s + (\delta s)^2 \approx 2 s \, \delta s$ when $\delta s$ is small compared with $s$.  So in terms of relative changes $\dfrac{\delta s}{\delta x} \approx \dfrac{1}{2s}$.  Throughout this process $\delta s$ and $\delta x$ are small
You then take the limit, substitute back  and get the derivative $\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2\sqrt{x}}$
A: When talking about derivatives here, we will be studying the variation of a certain function($x^2$, $\sqrt{x}$, ...) with respect to the variable $x$. In other words, we will be studying how the function changes, according to a very small variation $dx$ on $x$.
That is $\frac{df(x)}{dx}$ or in your examples $\frac{dx^2}{dx}$ and $\frac{d\sqrt{x}}{dx}$.
It is not a rule to divide what you call  the "difference in area" by the "small difference" to get the derivative. If you put $\frac{dx}{d\sqrt{x}}$ you'll be studying the variation of $x$ according to a small variation of $\sqrt{x}$ and thats not the definition of the derivative you're talking about.
