TITLE: Derivate by $x^2$
QUESTION [7 upvotes]: $$\frac{d}{dx^2}x=\frac{1}{2x}$$
I can't wrap my head around that. It appeared in a physics context, i.e. it may not be mathematically rigorous.
Thanks for some explanation.
Wolfram Alpha tells me I cannot differentiate by $x^2$: Invalid value

REPLY [2 votes]: $$\large \frac{df(x)}{dg(x)}=\large \frac{\frac{df(x)}{dx}}{\frac{dg(x)}{dx}}=\frac{f'(x)}{g'(x)}$$ for example $$f(x)=x^4+x^2+1 ,g(x)=x^2 $$
$$\frac{df}{dg}=\frac{d(x^4+x^2+1)}{d(x^2)}=\frac{4x^3+2x}{2x}=2x^2+1$$ if you take $u=x^2$ you need to find $\frac{df}{du}$ so first rewrite $f(x)$ by $u$ 
$$f(x)=(x^2)^2+x^2+1=u^2+u+1 \\ \to \frac{df}{du}=2u+1=2x^2+1$$ 
another example :$$f(x)=sinx ,g(x)=tan x \\\frac{df}{dg}=\frac{cos x}{tan^2 x+1}=sec^3x$$
another example : $$\frac{dx}{d(lnx)}=?\\=\frac{1}{\frac{1}{x}}=x$$