Derivative with respect to $x^2$

Question

What is the best notation for the first order as well as higher order derivative wrt $x^2$? I am using $$\frac{d}{dx^2}, \frac{d^2}{dx^4}, \frac{d^3}{dx^6}$$ and so on for the first, second and third order derivatives. Please let me know whether this notation is correct or confusing. Thanks in advance.

Answer 1

You have made a mistake ;

$\frac{d^2}{d(x^2)^2}\ne \frac{d^2}{dx^4} $

the first one implies the second derivative w.r.t $x^2$ while i dont even know what the latter represents.

Remember that the powers of $\frac{d^n}{da^n}$ cannot be multipied within ;

In your case when differentiating w.r.t $x^2$

the first , second and third derivatives and in general $n^{th}$ derivatives are; $\frac{d}{dx^2},\frac{d^2}{d(x^2)^2},\frac{d^3}{d(x^2)^3}\cdots\frac{d^n}{d(x^2)^n}$

Try keeping the variable with you are differentiating w.r.t inside brackets to avoid confusion . Or you can use a change of variables and set it equal to $u$ and write it in terms of $u$

Answer 2

Personally I would use $$\frac{d}{d(x^2)}, \frac{d^2}{d(x^2)^2}, \frac{d^3}{d(x^2)^3} $$

As I think this is clearer

Answer 3

In $$\frac{d^2}{dx^2}$$ the denominator denotes the square of the differential (and not a second order differential). It should actually be denoted

$$\frac{d^2}{(dx)^2}.$$

So for the second order differentiation on $x^2$,

$$\dfrac{d^2}{(d(x^2))^2}.$$

Of course

$$\dfrac{d^2}{{dx^2}^2}$$ is dubious, while

$$\dfrac{d^2}{d(x^2)^2}\text{ or }\dfrac{d^2}{dx^4}$$ might be taken for

$$\dfrac{d^2}{d(x^4)}$$

(a second order differential) and $$\dfrac{d^2}{(dx^2)^2}\text{ or }\dfrac{d^2}{dx^4}\text{ (again)}$$ might be taken for

$$\dfrac{d^2}{(dx)^4},$$ which is meaningless.

Answer 4

Note: $$\frac{d}{dx}; \frac{d}{dx}\left(\frac{d}{dx}\right)=\frac{d^2}{dxdx}=\frac{d^{\color{red}{2}}}{(dx)^{\color{red}{2}}}=\frac{d^{\color{red}{2}}}{dx^{\color{red}{2}}};... \\ \frac{d}{dx^2}; \frac{d}{dx^2}\left(\frac{d}{dx^2}\right)=\frac{d^2}{dx^2dx^2}=\frac{d^{\color{red}{2}}}{(dx^2)^{\color{red}{2}}};...$$