You only need $\frac{d}{dx}$, $\frac{dy}{dx}$, and $f^\prime$.
$\frac{d}{dx}\Psi$, where $\Psi$ is an expression whose value depends on that of $x$, is the rate of change of the value of $\Psi$ with respect to $x$.
$$\frac{d}{dx}\Psi = \lim_{h \rightarrow 0} \frac{f\left(x+h\right) - f\left(x\right)}{h}\ \text{where}\ f = k \mapsto \left(\Psi,\ \text{given}\ x=k\right)$$
$\frac{dy}{dx}$ is the rate of change of a dependent variable ($y$) with respect to an independent variable ($x$). Even though $y$ is simply an expression whose value depends on $x$ (making $\frac{dy}{dx}$ and $\frac{d}{dx}y$ synonyms), $\frac{dy}{dx}$ is nice because its form reminds us of the intuition that comes from $\frac{\Delta y}{\Delta x}$.
$$\frac{dy}{dx}= \lim_{h \rightarrow 0} \frac{f\left(x+h\right) - f\left(x\right)}{h}\ \text{where}\ f = k \mapsto \left(y,\ \text{given}\ x=k\right)$$
$f^\prime$, where $f$ is a function with one argument, is the rate of change of change of $f$'s output with respect to $f$'s input.
$$f^\prime = k \mapsto \lim_{h \rightarrow 0} \frac{f\left(k+h\right) - f\left(k\right)}{h}$$
Some caveats:
$\frac{df}{dx}$ and $\frac{d}{dx} f$ are meaningless because $f$ is a function, not a variable or an expression. The value of $f$ does not change with $x$; the value of $f\left(x\right)$ changes with $x$. So use $\frac{d}{dx} f\left(x\right)$ instead. Note that $\frac{d}{dx} f\left(x\right)$ is a synonym of $f^\prime\left(x\right)$. Also note that $\frac{d}{dx} f\left(a\right)=0$ because $f\left(a\right)$ is not an expression that depends on $x$.
Likewise, $y^\prime$ is meaningless because $y$ is a variable, not a function. $y$ could depend on any number of variables, so there's no unambiguous way to write $y$ as a function. But you could theoretically write: $\left( k \mapsto \left(y,\ \text{given}\ x=k \right)\right)^\prime\left(x\right)$ which is a synonym of $\frac{dy}{dx}$.