Derivative of Cot x

The derivative of cot x is -1 times the square of csc x. Before this, let us recall some facts about cot x. Cot x (cotangent x) in a right-angled triangle is the ratio of the adjacent side of x to the opposite side of x and thus it can be written as (cos x)/(sin x). We use this in doing the differentiation of cot x.

Let us learn the derivative of cot x formula along with its proof (in different methods) and a few solved examples.

The derivative of cot x with respect to x is represented by d/dx (cot x) (or) (cot x)' and its value is equal to -csc²x. Cot x is a differentiable function in its domain. To prove the differentiation of cot x to be -csc²x, we use the trigonometric formulas and the rules of differentiation. We are going to prove this formula in the following ways:

• Proof by first principle
• Proof by chain rule
• Proof by quotient rule

Derivative of cot x Formula

The formula for differentiation of cot x is,

• d/dx (cot x) = -csc²x (or)
• (cot x)' = -csc²x

Let us prove this in each of the above mentioned methods.

To find the derivative of cot x by first principle, we assume that f(x) = cot x. By the first principle (or definition of derivative), the derivative of f(x) is given by the following limit.

f'(x) = limₕ→₀ [f(x + h) - f(x)] / h ... (1)

Since f(x) = cot x, we have f(x + h) = cot (x + h).

Substituting these in (1),

f'(x) = limₕ→₀ [cot(x + h) - cot x] / h

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

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

By sum and difference formulas, sin A cos B - cos A sin B = sin (A - B).

f'(x) = limₕ→₀ [ sin (x - (x + h)) ] / [ h sin x · sin(x + h)]

= limₕ→₀ [ sin (-h) ] / [ h sin x · sin(x + h)]

We have sin (-h) = - sin h.

f'(x) = - limₕ→₀ (sin h)/ h · limₕ→₀ 1 / [sin x · sin(x + h)]

By limit formulas, limₕ→₀ (sin h)/ h = 1.

f'(x) = -1 [ 1 / (sin x · sin(x + 0))] = -1/sin²x

We know that the reciprocal of sin is csc. So,

f'(x) = -csc²x.

Hence proved.

We can prove the derivative of cot x formula by chain rule. For this, let us recall that cot and tan are reciprocals of each other. So we can write y = cot x as 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. Also, one of the properties of exponents says, a^-m = 1/a^m.

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)

= -1/sin²x

We know that the reciprocal of sin is csc. i.e., 1/sin x = csc x. So

y' = -csc²x

Hence proved.

We will find the derivative of cot x using quotient rule. This proof is the easiest among all the other methods of proving the derivative of cot x. Since the quotient rule deals with derivatives of fractions, we will write cot x as a fraction. We know that cot x = (cos x)/(sin x). So we assume that y = (cos x)/(sin x). Then by quotient rule,

y' = [ sin x · d/dx (cos x) - cos x · d/dx (sin x)] / (sin²x)

= [sin x · (- sin x) - cos x (cos x)] / (sin²x)

= [-sin²x - cos²x] / (sin²x)

= -[sin²x + cos²x] / (sin²x)

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

y' = -1 / (sin²x) = -csc²x

Hence proved.

Common Misconceptions Related to Derivative of Cot x:

Here are some common misconceptions related to the differentiation of cot x. Let us clarify.

• Though cot x = (cos x)/(sin x), d/dx (cot x) is NOT equal to d/dx(cos x) / d/dx (sin x).
  We have to use the quotient rule to find the derivative of cot x (by writing it as (cos x)/(sin x)).
• d/dx (cot x) is NOT tan x. Cot and tan are reciprocals of each other.
• The derivative of cot x and the derivative of cot inverse x are NOT same.
  d/dx(cot x) = -csc²x
  d/dx(cot^-1x) = -1/(1 + x²)

Topics Related to Differentiation of Cot x:

Here are some topics that you may find helpful while learning the derivative of cot x.

• Calculus Formulas
• Limit Formulas
• Derivative Formulas
• Calculus Calculator
• Derivative Calculator

What is the Derivative of Cot x with Respect to x?

The derivative of cot x with respect to x is -1 times the square of csc x. i.e., d/dx(cot x) = -csc²x. It can also be written as (cot x)' = -csc²x.

How to Prove the Derivative of Cot x Formula?

Let y = cot x. We have cot x = sin x/cos x. By quotient rule, y' = [ sin x · d/dx (cos x) - cos x · d/dx (sin x)] / (sin²x) = [-sin²x - cos²x] / (sin²x) = -1 / (sin²x) = -csc²x.

What is the Derivative of Cot x^2?

We know that d/dx(cot x) = -csc²x. So d/dx(cot x²) = -csc²x² d/dx(x²) = -2x csc²x² (by chain rule).

What is the Differentiation of Cot x in Terms of Sin x?

We know that the derivative of cot x is -csc²x. Also, csc x = 1/(sin x). So d/dx (cot x) = -1/sin²x.

What is the Derivative of Cot x^-1?

By using the derivative of cot x and chain rule, d/dx(cot x^-1) = -csc²x^-1 d/dx(x^-1) = -csc²x^-1 (-1 x^-2) = (csc²x^-1)/(x²).

Is the Derivative of Cot x Equal to Derivative of Cot Inverse x?

No, the derivatives of cot x and cot^-1x are different. The derivative of cot x is -csc²x whereas the derivative of cot⁻¹x is -1/(1 + x²).

What is the Difference Between the Derivative of Cot x and the Antiderivative of Cot x?

The derivative of cot x is -csc²x. The antiderivative of cot x is nothing but the integral of cot x and ∫ cot x dx = ln |sin x| + C.

What is the Derivative of Cot x²?

We know that d/dx(cot x) = -csc²x. So d/dx(cot x³) = -csc²x³ d/dx(x³) = -3x² csc²x² (by chain rule).