Square of a second derivative is the fourth derivative

I have a simple question for you guys, if I have this:

$$\left(\frac{d^2}{{dx}^2}\right)^2$$

Is it equal to this:

$$\frac{d^4}{{dx}^4}$$

Such that if I have an arbitrary function $f(x)$ I can get:

$$\left(\frac{d^2 f(x)}{{dx}^2}\right)^2 = \frac{d^4 f(x)}{{dx}^4}$$

Sorry if it's a pretty simple question, but I was trying to simplify something like this:

$$\left(a \cdot \frac{d^2}{{dx}^2} - f(x)\right)^2 g(x)$$ that's where it came up.

It depends on what you mean by "square" and it's basically a problem of notation. When you write $\left(\frac{d^2}{{dx}^2}\right)^2 $, implicitly the "square" means that you compose the operator $\frac{d^2}{{dx}^2}$ with itself, i.e. you consider $\frac{d^2}{{dx}^2} \circ \frac{d^2}{{dx}^2}$. This is of course equal to $\frac{d^4}{{dx}^4}$: differentiating four times is the same thing as differentiating twice then differentiating twice again. Applied to some function $f$, this then gives $\frac{d^2}{{dx}^2} \left( \frac{d^2 f(x)}{{dx}^2} \right) = \frac{d^4 f(x)}{{dx}^4}$, which is true.

On the other hand, where you write $\left(\frac{d^2 f(x)}{{dx}^2}\right)^2$, the "square" is implicitly multiplication, i.e. you're considering $\left(\frac{d^2 f(x)}{{dx}^2}\right) \cdot \left(\frac{d^2 f(x)}{{dx}^2}\right)$. This is not equal to the first thing, as simple counterexamples show (e.g. $f(x) = x^3$: $f''(x) = 6x$ thus $(d^4f)/(dx^4)(x) = 0$ while $((d^2f)/(dx^2)(x))^2 = (6x)^2 = 36x^2$). So you need to be careful with your notations.

I'm afraid not. Consider calculating the two quantities for

$$f(x)=x^3.$$

No. If $f(x) = x^4$ then $\frac{d^4f}{dx^4} = 24$ whereas $\left(\frac{d^2f}{dx^2}\right)^2 = (12x^2)^2 = 144x^4$.

It turns out that the first fact you cited actually does apply. The difficulty appears to be confusion between the expressions $\left(a \cdot \frac{d^2}{{dx}^2} - f(x)\right)^2$ and $\left(a \cdot \frac{d^2}{{dx}^2} f(x)\right)^2$

The expression in parentheses has two terms, and expands like this:

\begin{align} \left(a \frac{d^2}{{dx}^2} - f(x)\right)^2 g(x) &= \left(a \frac{d^2}{{dx}^2} - f(x)\right) \left(a \frac{d^2}{{dx}^2} - f(x)\right) g(x) \\ &= \left(a \frac{d^2}{{dx}^2} - f(x)\right) \left(a \frac{d^2}{{dx}^2} g(x)- f(x)g(x)\right) \\ &= a \frac{d^2}{{dx}^2} \left(a \frac{d^2}{{dx}^2} g(x)- f(x)g(x)\right) - f(x)\left(a \frac{d^2}{{dx}^2} g(x)- f(x)g(x)\right) \\ &= a^2 \frac{d^4}{{dx}^4} g(x) - a \frac{d^2}{{dx}^2} \left(f(x)g(x)\right) - a f(x) \frac{d^2}{{dx}^2} g(x) + (f(x))^2 g(x) \end{align}