$\dfrac{d}{dx} \, x = 1$

The derivative of $x$ with respect to $x$ can be derived in mathematics by using the fundamental form of differentiation of a function in limits.

$$\dfrac{d}{dx} \, f(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$$

Take $f(x) = x$, then $f(x+h) = x+h$. Substitute them in limits formula to derive the derivative of $x$ with respect to $x$.

$$\implies \dfrac{d}{dx} \, x = \lim_{h \to 0} \dfrac{x+h-x}{h}$$

The numerator contains two $x$ terms with opposite signs. So, they both cancelled each other as per the subtraction rule of literals.

$$\implies \require{cancel} \dfrac{d}{dx} \, x = \lim_{h \to 0} \dfrac{\cancel{x}+h-\cancel{x}}{h}$$

$$\implies \dfrac{d}{dx} \, x = \lim_{h \to 0} \dfrac{h}{h}$$

The numerator and denominator have same literals and the quotient of them is one by the division rule of literals.

$$\implies \require{cancel} \dfrac{d}{dx} \, x = \lim_{h \to 0} \dfrac{\cancel{h}}{\cancel{h}}$$

$$\implies \dfrac{d}{dx} \, x = \lim_{h \to 0} 1$$

There is no $h$ term in the limits function. So, the value limit $h$ tends to zero for $1$ is one.

$\therefore \,\,\,\,\,\, \dfrac{d}{dx} \, x = 1$

Therefore, it is proved that the derivative of $x$ with respect to $x$ is one.

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