Derivative of Cot x

Derivative of Cot x is -cosec²x. It refers to the process of finding the change in the sine function with respect to the independent variable. Derivative of cot x is also known as differentiation of cot x which is the process of finding rate of change in the cot trigonometric function.

In this article, we will learn about the derivative of cot x and its formula including the proof of the formula using the first principle of derivatives, quotient rule, and chain rule as well.

What is Derivative of Cot x?

The derivative of cot x is -cosec²x. The derivative of cot x is one of the six trigonometric derivatives that we have to study. It is the differentiation of trigonometric function cotangent with respect to the variable x in the present case. If we have cot y or cot θ then we differentiate the cotangent with respect to y or θ respectively.

Learn, Derivative in Math

Derivative of Cot x Formula

The formula of the derivative of cot x is given by:

(d/dx)[cot x] = -cosec²x

or

(cot x)’ = -cosec²x

Proof of Derivative of Cot x

The derivative of cot x can be proved using the following ways:

By using First Principle of Derivative
By using Quotient Rule
By using Chain Rule

Derivative of Cot x by First Principle of Derivative

Let’s start the proof for derivative of Cot x:

Let f(x) = Cot x

By the First Principle of Derivative

f'(x)= lim h→0 f(x+h)-f(x)/h

= lim h→0 cot(x+ h)- cot x/ h

= lim h→0 [cos(x+h)/sin(x+h)- cos x/ sin x]/h

= lim h→0 sin x cos(x+h)-cos x sin (x+h) / sin(x+h) sin x. h

=lim h→0 sin [x-(x+h) / sin(x+h).sin x .h

= lim h→0 – sin h/h lim h→0 1/sin (x+h)sin x

= -1 × 1/sinx. sinx

= -1/ sin²x

= -cosec²x

Derivative of Cot x by Quotient Rule

To find the derivative of cot x using the quotient rule of derivative we have to use the following mentioned formulas

(d/dx) [u/v] = [u’v – uv’]/v²
sin²(x)+ cos²(x)= 1
cot x = cos x / sin x
cosec x = 1 / sin x

Let’s start the proof of the derivative of cot x

f(x) = cot x = cos(x)/sin(x)

u(x) = cos(x) and v(x)=sin(x)

u'(x) = -sin(x) and v'(x)=cos(x)

v²(x) = sin²(x)

f'(x) = {-sin(x).sin(x) – cos(x).cos(x)}/sin²(x)

f'(x) = -sin²(x)-cos²(x)/sin²(x)

f'(x) = -sin²(x)+cos²(x)/sin²(x)

By one of the trigonometric identities, cos²x + sin²x = 1.

f'(x) = – 1/ sin²(x)

d/dx cot(x) = -1 /sin²(x) = -cosec²(x)

Therefore, differentiation of cot x is -cosec²x.

Derivative of Cot x by Chain Rule

Assume y = cot x then we can write y = 1 / (tan x) = (tan x)^-1. Since we have power here, we can apply the power rule here. By power rule and chain rule,

y’ = (-1) (tan x)^-2·d/dx (tan x)

The derivative of tan x is, d/dx (tan x) = sec²x

y= cot x

y’ = -1/tan²x·(sec²x)

y’ = – cot²x·sec²x

Now, cot x = (cos x)/(sin x) and sec x = 1/(cos x). So

y’ = -(cos²x)/(sin²x) · (1/cos²x)

y’ = -1/sin²x

Since, reciprocal of sin is cosec. i.e., 1/sin x = cosec x. So

y’ = -cosec²x

Hence proved.

Also Read,

Differentiation of Trigonometric Function

Differentiation Formulas

Solved Examples on Derivative of Cot x

Some examples related to Derivative of Cot x are,

Example 1: Find the derivative of cot²x.

Solution:

Let f(x) = cot²x = (cot x)²

By using power rule and chain rule,

f'(x) = 2 cot x · d/dx(cot x)

We know that the derivative of cot x is -cosec²x. So

f'(x) = -2 cot x ·cosec²x

Example 2: Differentiate tan x with respect to cot x.

Solution:

Let v = tan x and u = cot x. Then dv/dx = sec²x and du/dx = -cosec²x.

We have to find dv/du. We can write this as

dv/du = (dv/dx) / (du/dx)

dv/du = (sec²x) / (-cosec²x)

dv/du = (1/cos²x) / (-1/sin²x)

dv/du = (-sin²x) / (cos²x)

dv/du = -tan²x

Example 3: Find the derivative of cot x · csc2x

Solution:

Let f(x) = cot x · cosec²x

By product rule,

f'(x) = cot x·d/dx (cosec²x) + cosec²x·d/dx(cot x)

f'(x) = cot x·(2 cosec x) d/dx (cosec x) + cosec²x (-cosec²x) (by chain rule)

f'(x) = 2 cosec x cot x (-cosec x cot x) – cosec⁴x

f'(x) = -2 cosec²x cot²x – cosec⁴x