\( \cot x \) derivative is -1 times the square of \( \csc \) Let us first review some facts about cot x. In a right-angled triangle, cot x (cotangent x) is the ratio of the adjacent side of x to the opposite side of x, and can thus be written as (cos x)/ (sin x). This is used in the differentiation of cot x.

One of the first transcendental functions introduced in Differential Calculus is the Derivative of Cotangent (or Calculus I).

Let us go over the derivative of cot x formula, as well as its proof (in various methods) and a few solved examples.

What is Cotx?

Cot(x) is a trigonometric function that represents the ratio of the adjacent side to the opposite side of a right triangle where the angle opposite the adjacent side is x. More specifically, cot(x) is defined as the reciprocal of the tangent function, or cot(x) = 1/tan(x).

Alternatively, it can be expressed in terms of the cosine and sine functions as cot(x) = cos(x)/sin(x). 

The cotangent function is periodic with a period of π, and it has singularities at the zeros of the sine function, which correspond to the vertical asymptotes of the graph. The range of cot(x) is all real numbers, except for 0, which is not in the range since sin(x) and cos(x) cannot be zero simultaneously.

Derivative of cot x

If x is used to represent a variable, the cotangent function is written as cot x in mathematics.

The differentiation of the cot function with respect to x is written in differential calculus in the following mathematical form.

\( \frac{\mathrm{d} }{\mathrm{d} x} \left ( \cot x \right ) = -\csc ^{2}x \)

Derivative of Cotx Formula

The derivative of cot(x) is given by the formula d/dx(cot(x)) = -csc^2(x), where "csc" stands for cosecant. This formula represents the rate of change of the cotangent function with respect to its input x. It states that the slope of cot(x) at any given point x is equal to the negative cosecant squared of that point. 

The derivative of cot(x) is an important concept in calculus, and it is used to solve various problems involving the rate of change, optimization, and integration. 

Derivative of Cot x Proof by First Principle

We assume that \( f\left ( x \right ) = \cot x \) in order to find the derivative of cot x using first principles. The derivative of f(x) is given by the following limit according to the first principle (or definition of derivative).

Step 1: \( f’ \left ( x \right ) = \lim_{h\rightarrow 0}\frac{\left [ f\left ( x+h \right ) \right ]}{h} \) ….(1)

And,

Step 2: \( f\left ( x \right )= \cot x \)

Step 3: \( f\left ( x +h \right )= \cot \left ( x+h \right ) \)

In equation (1) substract these values

Step 4: \( = \lim_{h\rightarrow 0}\left [ \frac{\sin \left ( -h \right )}{h\sin x.\sin \left ( \left ( x+h \right ) \right )}\right ]\)

Step 5: \( \frac{\cos x}{\sin x}{h}\)

Step 6: \( = \lim_{h\rightarrow 0}\left [ \frac{\frac{\sin x\cos x\left ( x+h \right )-\cos x\sin \left ( x+h \right)}{\sin x.\sin \left ( x+h \right )}}{h} \right ] \)

Using the sum and difference formulas,

Step 7: \( \sin A\cos B -\cos A\sin B = \sin \left ( A-B \right ) \)

Step 8: \( f’\left ( x \right ) = \lim_{h\rightarrow0 } \left [ \frac{\sin \left ( x-\left ( x+h \right ) \right )}{h\sin x.\sin \left ( x+h \right )} \right ] \)

Step 9: \( = \lim_{h\rightarrow0 } \left [ \frac{\sin \left ( -h \right )}{h\sin x.\sin \left ( \left ( x+h \right ) \right )}\right ] \)

We have,

Step 10: \( \sin \left ( -h \right ) = -\sin h \)

Step 11: \( f’\left ( x \right ) = -\lim_{h\rightarrow 0} \frac{\left ( \sin h \right )}{h}.\frac{\lim_{h\rightarrow0}}{\left [ \sin x.\sin \left ( x+h \right ) \right ]} \)

Step 12: \( \frac{\lim_{h\rightarrow 0}\left ( \sin h \right ) }{h} =1 \)

Step 13: \( f’\left ( x \right )=-1\left [ \frac{1}{\left ( \sin x.\sin \left ( x+0 \right ) \right )} \right ] = \frac{-1}{\sin^{2}x} \)

We know that sin reciprocal is csc. So,

\( f’\left ( x \right )=- – \csc ^{2}x \)

Hence proved

Derivative of Cot x Proof by Chain Rule

We can use the chain rule to prove the derivative of the cot x formula. Let us remember that cot and tan are reciprocals of each other.

The power rule can be applied here. According to the power and chain rules,

Step 1: \( y’= \left ( -1 \right )\left ( \tan x \right )^{-2} . \frac{\mathrm{d} }{\mathrm{d} x} \left ( \tan x\right ) \)

Step 2: \( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \tan x \right ) = \sec ^{2}x \)

Step 3: \( y’ = \frac{1}{\tan ^{2}x . \left ( \sec ^{2}x \right )} \)

Step 4: \( y’ = – \cot ^{2}x. \sec ^{2}x \)

Step 5: \( \cot x = \frac{\cos x}{\sin x} and \sec x = \frac{1}{\left ( \cos x \right )} \)

Step 6: \( y’ = \frac{\left ( \cos ^{2}x \right )}{\left ( \sin ^{2}x\right ). \left ( \frac{1}{\cos ^{2}x} \right )} \)

Step 7: \( = \frac{-1}{\sin ^{2}x} \)

We know that sin’s reciprocal is csc. \( \frac{1}{\sin x} = \csc x \)

\( y’ = – \csc ^{2}x \)

Hence proved

Derivative of Cot x Proof by Quotient Rule

Step 1: \( \frac{\mathrm{d} }{\mathrm{d} x} \frac{u}{v} =\frac{vu’-uv’}{v^{2}} \)

Step 2: \( \cot x = \frac{\cos x }{\sin x} \) apply Quotient Rule

Step 3: \( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}\cot x \)

Step 4: \( \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \frac{\cos x}{\sin x}\)

Step 5: \( \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\sin x.\left ( -\sin x \right ) – \cos x. \cos x}{\sin ^{2}x} \)

Step 6: \( \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-\sin ^{2}x+\cos ^{2}x}{\sin ^{2}x} \)

Step 7: \( \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = -\frac{1}{\sin ^{2}x} \)

Step 8: \( \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = – \csc ^{2}x \)

Hence proved

Derivative of cot x Solved Examples

Problem: 1
What is the derivative of tan x in terms of cot x?

Solution:

\( v = \tan x \)

\( u = \cot x \)

\( \frac{\mathrm{d} v}{\mathrm{d} x} = \sec ^{^{2}}x \)

\( \frac{\mathrm{d} u}{\mathrm{d} x} = \csc ^{^{2}}x \)

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

\( \frac{\mathrm{d} v}{\mathrm{d} u} = \frac{\left ( \frac{\mathrm{d} v}{\mathrm{d} x} \right )}{\left ( \frac{\mathrm{d} u}{\mathrm{d} x} \right )} \)

\( = \frac{\left ( \sec ^{2}x \right )}{- \csc ^{2}x} \)

\( = \frac{\left ( \frac{1}{\cos ^{2}x} \right )}{\left ( \frac{-1}{\sin ^{2}x} \right )} \(\) \(\) = – \tan ^{2}x \)

Tan x’s derivative with respect to cot x is \( = – \tan ^{2}x \)

Problem: 2
Determine the derivative of \( \cot x. \csc ^{2}x \)

Solution:

Let,

\( f\left ( x \right )=\cot x. \csc ^{2}x \)

Using product rule,

\( f\left ( x \right )=\cot x.\frac{\mathrm{d} }{\mathrm{d} x}\left ( \csc ^{2}x \right ) + \csc ^{2}x\frac{\mathrm{d} }{\mathrm{d} x}\left ( \cot x \right ) \)

using chain rule

\( = \cot x. \left ( 2\csc x \right )\frac{\mathrm{d} }{\mathrm{d} x}\left ( \csc x \right ) + \csc ^{2}x\left ( -\csc ^{2}x\right ) \)

\( = 2 \csc x \cot x\left ( -\csc x \cot x \right ) – \csc ^{4}x \)

\( = – 2 \csc ^{2}x \cot ^{2}x – \csc ^{4}x \)

The derivative of \( \cot x. \csc ^{2}x \) is \( = – 2 \csc ^{2}x \cot ^{2}x – \csc ^{4}x \)

Problem 3: Find the derivative of 1/cotx

Solution:

We rewrite 1/cot x as tan x because cot x = 1/tan x. Therefore, 1/cot x = tan x.

Now, we can use the derivative formula for tan x which is sec^2 x. Therefore:

\(\begin{align*}\frac{d}{dx}\left(\frac{1}{\cot x}\right) &= \frac{d}{dx}(\tan x) \&= \frac{d}{dx}\left(\frac{1}{\cot x}\right) \&= \frac{d}{dx}\left(\frac{1}{\frac{1}{\tan x}}\right) \&= \frac{d}{dx}(\tan x) \&= \sec^2 x\end{align*}\)

So the derivative of 1/cot x with respect to x is sec^2 x.

Problem 4: Find the derivative of cotx sinx

Solution:  We can start by using the product rule of differentiation, which states that for two functions $u(x)$ and $v(x)$, the derivative of their product $u(x)v(x)$ is given by:

(u(x) v(x))′=u′(x) v(x) + u(x) v′(x)

(cotx sinx)′= (cotx)′ Sinx + cotx (sinx)′

Applying this rule to cot x sin x, we get:

We can simplify this expression by finding the derivatives of $\cot x$ and $\sin x$. The derivative of $\cot x$ is $-\csc^2 x$, and the derivative of $\sin x$ is $\cos x$. Therefore:

\(\begin{align*}(\cot x \sin x)' &= (-\csc^2 x)\sin x + \cot x (\cos x) \&= -\frac{\sin x}{\sin^2 x} + \frac{\cos x}{\sin x} \&= -\frac{1}{\sin x} + \frac{\cos x}{\sin x}

\end{align*}\)

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