Derivative of Cosec x (original) (raw)

Last Updated : 23 Jul, 2025

**Derivative of Cosec x is **-Cot x Cosec x. The derivative of cosec x is represented by the d/dy(cosec x). It explains about the slope of the graph of cosec x. Cosecant Functions are denoted as csc or cosec and defined as the reciprocal of the sine function i.e., 1/sin x.

In this article, we will discuss all the topics related to the derivative of cosec x including its proof using various methods. Let’s start our learning on the topic of Derivative of Cosec x.

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What is Derivative of Cosec x?

Among the trig derivatives, the derivative of the cosec x is one of the derivatives. The derivative of the cosec x is -cot x cosec x. The derivative of cosec x is the rate of change with respect to the angle i.e., x. The resultant of the derivative of cosec x is -cot x cosec x.

Derivative of Cosec x Formula

The formula for the derivative of cosec x is given by:

****(d/dx) [cosec x] = -cot x × cosec x**

****(cosec x)’ = -cot x × cosec x**

Before moving forward we must learn about Derivative in Maths.

What is Derivative in Math?

Derivative of a function is the rate of change of the function with respect to any independent variable. The derivative of a function f(x) is denoted as **f'(x) or ****(d /dx)[f(x)]**.

The differentiation of a trigonometric function is called a derivative of the trigonometric function or trig derivatives.

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Proof of Derivative of Cosec x

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

Derivative of Cosec x by First Principle of Derivative

To prove derivative of cosec x using First Principle of Derivative, we will use basic limits and trigonometric formulas which are listed below:

Let’s start the proof for the derivative of cosec x

By First Principle of Derivative

Let y = cosec x

y = 1/sin x

⇒ y’ = d/dx (1/sin x)

⇒ y’ = lim h→0 (1/sin(x + h) - 1/sin x) / ((x + h) - x)

⇒ y’ = lim h→0 ((sin x - sin(x + h)) / (sin x × sin(x + h))) / h

⇒ y’ = lim h→0 (sin x - sin(x + h)) / (h × sin x × sin(x + h))

⇒ y’ = lim h→0 - (sin(x + h) - sin x) / (h × sin x × sin(x + h))

⇒ y’ = lim h→0 - (sin(x + h) - sin x) /h × lim h→0 1 /(sin x × sin(x + h))

⇒ y’ = -cos x × 1 / sin2 x

⇒ y’ = -cos x / sin x × 1 / sin x

⇒ y’ = -cot x × cosec x

Therefore, the differentiation of cosec x is – cosec x cot x.

Derivative of Cosec x by Quotient Rule

To prove the derivative of cosec x using the Quotient rule, we will use basic derivatives and trigonometric formulas which are listed below:

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

y = cosec x

⇒ y = 1/sin x

⇒ y’ = d/dx (1/sin x)

Applying quotient rule

y’ = ((d/dx) (1) × sin x – 1 × (d/dx)(sin x))/sin2 x

⇒ y’ = ((0) × sin x – (1) × (cos x))/sin2 x

⇒ y’ = -cos x/(sin x)2

⇒ y’ = -cot x × cosec x

Therefore, the differentiation of cosec x is – cosec x cot x.

Derivative of Cosec x by Chain Rule

To prove derivative of cosec x we will use chain rule and some basic trigonometric identities and limits formula. The trigonometric identities and limits formula which are used in the proof are given below:

Let’s start the proof for the differentiation of the trigonometric function cosec x

(d/dx) cosec x = (d/dx) (1 / sin x)

Using chain rule

(d/dx) cosec x = (-1 / sin2x) (d/dx) sin x

⇒ (d/dx) cosec x = (-1 / sin2x) cos x

⇒ (d/dx) cosec x = -(1 / sin x) (cos x / sin x)

⇒ (d/dx) cosec x = – cosec x cot x

Therefore, the differentiation of cosec x is – cosec x cot x.

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Conclusion

In conclusion, the derivative of csc⁡x is -cot⁡x⋅csc⁡x, which indicates how the cosecant function changes with respect to x. We have learned several methods to derive this result which includes the First Principle of Derivative, Quotient Rule, and Chain Rule. Understanding these techniques helps in analyzing and solving problems involving trigonometric functions effectively.

Examples Using Derivative of Cosec x

Some examples on Using Derivative of Cosec x are,

**Example 1: Find the derivative of cosec 4x.

**Solution:

Let y = cosec 4x

y’ = (d/dx) [cosec 4x]

Applying chain rule

y’ = (d/dx) [cosec 4x].(d/dx) (4x)

⇒ y’ = (-cot 4x × cosec 4x) × 4

⇒ y’ = -4 × cot 4x × cosec 4x

**Example 2: Evaluate the derivative f(x) = (x 3 + 5x 2 + 2x + 7) × cosec x.

**Solution:

f(x) = (x3 + 5x2 + 2x + 7) × cosec x

⇒ f'(x) = (d /dx)[(x3 + 5x2 + 2x + 7) × cosec x]

Applying product rule

⇒ f'(x) = (d /dx)[(x3 + 5x2 + 2x + 7)] × cosec x + (x3 + 5x2 + 2x + 7) × (d /dx)[cosec x]

⇒ f'(x) = (3x2 + 10x + 2) × cosec x + (x3 + 5x2 + 2x + 7) × (-cot x × cosec x)

**Example 3: Determine the second derivative of cosec x.

**Solution:

The first derivative of cosec x is -cosec x cot x.

To determine the second derivative of cosec x, we differentiate -cosec x cot x using the product rule.

(cosec x)'' = (-cosec x cot x)'

⇒ (-cosec x)' cot x + (-cosec x) (cot x)'

⇒ cosec x cot x cot x + (-cosec x) (-cosec2x)

⇒ cosec x (cot2x + cosec2x)

Second derivative of cosec x is cosec x (cot2x + cosec2x).

**Example 4: Find the derivative of cosec -1 x.

**Solution:

d/dx[cosec-1 x] = -1 / (|x| × sqrt(x2 - 1)), from formula

**Example 5: Evaluate the derivative cosec 5x + x × cosec x.

**Solution:

Let z = cosec 5x + x × cosec x

Differentiating

z’ = (d/dx) [cosec 5x + x × cosec x]

⇒ z’ = (d/dx) cosec 5x + (d/dx)[x × cosec x]

Applying chain rule and product rule

z’ = -5 × cot 5x × cosec 5x + (d/dx)(x) × cosec x + x × (d/dx)(cosec x)

⇒ z’ = -5 × cot 5x × cosec 5x + cosec x + x × (-cot x × cosec x)

⇒ z’ = -5 × cot 5x × cosec 5x + cosec x - x × cot x × cosec x

Practice Problems on Derivative of Cosec x

**Q1: Find the derivative of cosec 7x.

**Q2: Find the derivative of x 2 × cosec x.

**Q3: Evaluate: (d/dx) [cosec x / (x 2 + 2)].

**Q4: Evaluate the derivative of: cosec x × cot x

**Q5: Find: (cot x) cosec x .

**Q6: Compute the derivative of the function h(x)=csc⁡ 2 (x).

**Q7: Find the derivative of f(x)=csc⁡(x) with respect to x.

**Q8: Given the function g(x)=csc⁡(x)/cot⁡(x) find its derivative.

**Q9: Find the derivative of f(x)=3csc⁡(x)−5cot⁡(x) with respect to x.

**Q10: Determine the derivative of the function g(x)=csc⁡(x)+cot⁡(x).