Second derivative of itself

Question

I know the $ \frac{d^2x}{dx^2}= 0 $, since $dx/dx = 1 ...$ But by playing with some equations it is easy to get that $d^2f/dx^2=f''(x)$, so $d^2f=f''(x)dx^2$ and $df=f'(x)dx$, so $df^2=f'(x)^2dx^2$. So what is: $$d^2f/df^2$$ It is easy, just substitute the expressions for $d^2f$ and $df^2$ you will get: $$d^2f/df^2=f''dx^2/f'^2dx^2=f''/f'^2$$ Can you spot the mistake?

Answer 1

Simple.

Both of the following are false:

$$\frac{d^2y}{dx^2}\equiv\bigg(\frac{dy}{dx}\bigg)^2$$

$$\frac{df}{dx}=g(x)\implies\frac{df^2}{dx}=g^2(x)$$

To counter the second one, we take $f(x)=x^k$ then $[f'(x)]^2=k^2x^{2k-2}$ while $[f^2(x)]'=2kx^{2k-1}$