The derivative of cot function with respect to a variable is equal to the negative of square of cosecant function. If x is taken as a variable and represents an angle of a right triangle, then the co-tangent function is written as cot x in mathematics. The differentiation of the cot x function with respect to x is equal to −csc^2 x or −cosec^ 2 x, and it can be proved from first principle in differential calculus.

## The image below contains the derivative of cot x

**\(\frac{\partial }{\partial x}\cot x= -\csc^{2}x\)**

\(\frac{\partial }{\partial x}\cot x= \frac{\partial }{\partial x}\left(\frac{1}{\tan x}\right)\),

\(\frac{\partial }{\partial x}\cot x= \frac{(\tan x)(0)-(1)(\sec^{2}x)}{\tan^{2} x}\),

\(\frac{\partial }{\partial x}\cot x= -\frac{\sec^{2}x}{\tan^{2} x} \) ,

\(\frac{\partial }{\partial x}\cot x=- {\sec^{2} x}.{ \cot^{2} x} \) ,

\(\frac{\partial }{\partial x}\cot x=-\left(\frac{1}{\cos^{2} x}, \right)\left(\frac{\cos^{2}x}{\sin^{2}x}\right) \) ,

**\(\frac{\partial }{\partial x}\cot x= -\csc^{2}x\)**

### This is the way to find the derivative of cot x.and there are many ways to prove derivation of cot x but this is the simplest way from our side

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