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

In inverse trigonometry, the inverse cotangent function is written as $\cot^{-1}{x}$ or $\operatorname{arccot}{(x)}$ when $x$ represents a variable and a real number. The derivative of cot inverse function is written in two mathematical forms in differential calculus.

$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\cot^{-1}{(x)}\Big)}$

$(2) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\operatorname{arccot}{(x)}\Big)}$

The differentiation of the inverse cot function with respect to $x$ is equal to the negative reciprocal of the sum of one and $x$ squared.

$\implies$ $\dfrac{d}{dx}{\, \Big(\cot^{-1}{(x)}\Big)}$ $\,=\,$ $-\dfrac{1}{1+x^2}$

The differentiation of cotangent inverse function can be written in any variable. The following examples are some examples to learn how to write the formula for the derivative of inverse cotangent function in differential calculus.

$(1) \,\,\,$ $\dfrac{d}{du}{\,\Big(\cot^{-1}{(u)}\Big)} \,=\, \dfrac{-1}{1+u^2}$

$(2) \,\,\,$ $\dfrac{d}{dy}{\,\Big(\cot^{-1}{(y)}\Big)} \,=\, \dfrac{-1}{1+y^2}$

$(3) \,\,\,$ $\dfrac{d}{dz}{\,\Big(\cot^{-1}{(z)}\Big)} \,=\, \dfrac{-1}{1+z^2}$

Learn how to derive the differentiation of the inverse cotangent function from first principle.

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