$x$ is an angle of a right angled triangle and then cotangent function is $\cot{x}$. The differentiation or derivative of $\cot{x}$ function with respect to $x$ is written as $\dfrac{d}{dx} \cot{x}$ in differential calculus.

The derivative (or) differentiation of $cot{x}$ function is derived in differential calculus by expressing derivative of $\cot{x}$ function in limit form.

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

Take $f(x) = \cot{x}$, then $f(x+h) = \cot{(x+h)}$ and substitute them in the fundamental principle of differentiation of the function in limit form.

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

According to quotient identity of cot function with sin and cos functions, each cot function can be written as the quotient of cosine by sine function.

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\dfrac{\cos{(x+h)}}{\sin{(x+h)}}-\dfrac{\cos{x}}{\sin{x}}}{h}$

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\dfrac{\cos{(x+h)}\sin{x}-\sin{(x+h)}\cos{x}}{\sin{(x+h)}\sin{x}}}{h}$

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\cos{(x+h)}\sin{x}-\sin{(x+h)}\cos{x}}{h\sin{(x+h)}\sin{x}}$

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{-[\sin{(x+h)}\cos{x}-\cos{(x+h)}\sin{x}]}{h\sin{(x+h)}\sin{x}}$

The trigonometric expression in numerator represents sine of compound angle identity. So, the entire expression can be minimized by writing it as sin of compound angle.

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{ -\sin(x+h-x)}{h\sin(x+h)\sin x}$

$=\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{-\sin(\cancel{x}+h-\cancel{x})}{h\sin(x+h)\sin x}$

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{-\sin h}{h\sin(x+h)\sin x}$

Part of the trigonometric expression is same as the limit sinx/x as x approaches 0 formula. So, divide the trigonometric expression as two multiplying factors for taking first step in simplifying it mathematically.

$=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{-\sin{h}}{h} \times \dfrac{1}{\sin{(x+h)}\sin{x}}$

The limit value belongs to both multiplying factors. So, it can be applied to both functions to find value of them.

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize -\dfrac{\sin{h}}{h}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{1}{\sin{(x+h)}\sin{x}}$

$=\,$ $\displaystyle -\large \lim_{h \to 0} \normalsize \dfrac{\sin{h}}{h}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{1}{\sin{(x+h)}\sin{x}}$

The first multiplying factor represents lim sinx/x as x approaches 0 formula. So, its value is equal to 1 and substitute h is equal to 0 to obtain the value of the second multiplying factor. The entire mathematical expression is derivative of cotx function with respect to x.

$\implies$ $\dfrac{d}{dx} \cot{x}$ $\,=\,$ $-1 \times \dfrac{1}{\sin{(x+0)}\sin{x}}$

$\implies$ $\dfrac{d}{dx} \cot{x}$ $\,=\,$ $-\dfrac{1}{\sin{x} \sin{x}}$

$\implies$ $\dfrac{d}{dx} \cot{x}$ $\,=\,$ $-\dfrac{1}{\sin^2{x}}$

As per reciprocal identity of sin function with cosecant function, the reciprocal of sin function can be written as cosecant function.

$\implies$ $\dfrac{d}{dx} \cot{x} \,=\, \dfrac{-1}{\sin^2 x}$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{d}{dx} \cot{x} \,=\, -\csc^2{x}$

It is also written as follows.

$\implies \dfrac{d}{dx} \cot{x} \,=\, -\operatorname{cosec}^2{x}$

Therefore, it is proved that the differentiation (or) derivative of $\cot{x}$ with respect to $x$ is negative of square of cosecant of angle $x$.

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