Derivative of a Function with respect to Derivative of that Function To start, assume that $w(x)$ is a function such that $w : \mathbb{R} \rightarrow \mathbb{R}$. What would $\frac{d w}{d \frac{d w}{d x}}$ be?
My intuition says that 
$$
\frac{d}{d x}\frac{d w}{d \frac{d w}{d x}} = \frac{d}{d \frac{d w}{d x}}\frac{d w}{d x} = 1.
$$
If this is true, then it is clear that $\frac{d w}{d \frac{d w}{d x}}$ is nonzero. This, of course relies on all of these derivatives being continuous as well as $w$ itself. Taking a rather wild stab in the dark, my next step is to say that 
$$
\frac{d w}{d \frac{d w}{d x}} = \frac{d}{d x} \int \frac{d w}{d \frac{d w}{d x}} dx = \int \frac{d}{d x}\frac{d w}{d \frac{d w}{d x}} dx = \int 1 dx = x + c.
$$
This, of course, feels really hacky. Maybe my thought process is correct, but I definitely feel like I'm shooting in the dark.
 A: It’s not right because
$$\frac{d}{dx}\frac{dw}{d\frac{dw}{dx}}\neq \frac{d}{d\frac{dw}{dx}}\frac{dw}{dx}=1$$
There is a example:
You know $\frac{dt^2}{dt}=2t$, let $t=x^2$, so $\frac{dx^4}{dx^2}=2x^2$, so $$\frac{d}{dx}\left(\frac{dx^4}{dx^2}\right)=\frac{d2x^2}{dx}=4x$$
After that
$$\frac{d}{dx^2}\left(\frac{dx^4}{dx}\right)=\frac{d4x^3}{dx^2}=\frac{d4t^{3/2}}{dt}=6\sqrt t=6x\quad(t=x^2)$$
So
$$\frac{d}{dx}\frac{dx^4}{dx^2}\neq\frac{d}{dx^2}\frac{dx^4}{dx}$$
And what would $\frac{dw}{d\frac{dw}{dx}}$ be?
For $w=w(x)$, we know$$w’(x)=\frac{dw}{dx},w’’(x)=\frac{d\left(\frac{dw}{dx}\right)}{dx}$$
So$$\frac{w’’(x)}{w’(x)}=\frac{d\left(\frac{dw}{dx}\right)}{dx}\frac{1}{\frac{dw}{dx}}= \frac{d\left(\frac{dw}{dx}\right)}{dw}$$
So$$\frac{dw}{d\frac{dw}{dx}}=\frac{w’(x)}{w’’(x)}\quad\left(=\frac{\frac{dw}{dx}}{\frac{d}{dx}\left(\frac{dw}{dx}\right)}=\frac{\frac{dw}{dx}}{\frac{d^2w}{dx^2}}\right)$$
(It’s the first day I use English to write an answer, if there are any wrong expression, please tell me, I am very appreciated.)
