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Last updated on September 6, 2025

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Derivative of csc(x)

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We use the derivative of csc(x), which is -csc(x)cot(x), to understand how the cosecant function changes in response to a slight change in x. Derivatives are crucial in various fields, including physics and engineering. We will now discuss the derivative of csc(x) in detail.

Derivative of csc(x) for Saudi Students
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What is the Derivative of csc(x)?

We now understand the derivative of csc(x).

It is commonly represented as d/dx (csc x) or (csc x)', and its value is -csc(x)cot(x).

The function csc(x) has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Cosecant Function: csc(x) = 1/sin(x).

Quotient Rule: Rule for differentiating csc(x) (since it consists of 1/sin(x)).

Cotangent Function: cot(x) = cos(x)/sin(x).

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Derivative of csc(x) Formula

The derivative of csc(x) can be denoted as d/dx (csc x) or (csc x)'.

The formula we use to differentiate csc(x) is: d/dx (csc x) = -csc(x)cot(x) (csc x)' = -csc(x)cot(x)

The formula applies to all x where sin(x) ≠ 0.

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Proofs of the Derivative of csc(x)

We can derive the derivative of csc(x) using proofs.

To show this, we will use the trigonometric identities along with the rules of differentiation.

There are several methods we use to prove this, such as:

 

By First Principle

Using Chain Rule

Using Product Rule

We will now demonstrate that the differentiation of csc(x) results in -csc(x)cot(x) using the above-mentioned methods:

 

By First Principle

The derivative of csc(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of csc(x) using the first principle, we will consider f(x) = csc(x).

Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

Given that f(x) = csc(x), we write f(x + h) = csc(x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [csc(x + h) - csc(x)] / h = limₕ→₀ [1/sin(x + h) - 1/sin(x)] / h = limₕ→₀ [(sin(x) - sin(x + h)) / (h sin(x) sin(x + h))]

Using the identity sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2), f'(x) = limₕ→₀ [-2 cos((2x + h)/2) sin(h/2)] / [h sin(x) sin(x + h)]

Applying limit formulas, limₕ→₀ sin(h/2)/(h/2) = 1, f'(x) = -cos(x) / sin²(x)

As csc(x) = 1/sin(x) and cot(x) = cos(x)/sin(x), we have, f'(x) = -csc(x)cot(x).

Hence, proved.

 

Using Chain Rule

To prove the differentiation of csc(x) using the chain rule,

We use the formula: csc(x) = 1/sin(x) Consider f(x) = 1 and g(x) = sin(x)

So we get, csc(x) = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1)

Let’s substitute f(x) = 1 and g(x) = sin(x) in equation (1), d/dx (csc x) = [0.sin(x) - 1.cos(x)] / sin²(x) = -cos(x) / sin²(x) Since csc(x) = 1/sin(x) and cot(x) = cos(x)/sin(x), we can write: d/dx(csc x) = -csc(x)cot(x)

 

Using Product Rule

We will now prove the derivative of csc(x) using the product rule.

The step-by-step process is demonstrated below:

Here, we use the formula, csc(x) = 1/sin(x) csc(x) = (sin(x))⁻¹

Given that, u = 1 and v = (sin(x))⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (1) = 0 (substitute u = 1)

Here we use the chain rule: v = (sin(x))⁻¹ = (sin(x))⁻¹ (substitute v = (sin(x))⁻¹) v' = -1(sin(x))⁻² · cos(x) v' = -cos(x)/sin²(x)

Again, use the product rule formula: d/dx (csc(x)) = u'.v + u.v'

Let’s substitute u = 1, u' = 0, v = (sin(x))⁻¹, and v' = -cos(x)/sin²(x)

When we simplify each term: We get, d/dx (csc(x)) = -cos(x)/sin²(x) = -csc(x)cot(x)

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Higher-Order Derivatives of csc(x)

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.

To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes.

Higher-order derivatives make it easier to understand functions like csc(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.

The second derivative is derived from the first derivative, which is denoted using f′′(x).

Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.

For the nth Derivative of csc(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change. (continuing for higher-order derivatives).

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Special Cases:

When x is 0, the derivative of csc(x) is undefined because csc(x) has a vertical asymptote there. When x is π/6, the derivative of csc(x) = -csc(π/6)cot(π/6), which is -2√3.

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Common Mistakes and How to Avoid Them in Derivatives of csc(x)

Students frequently make mistakes when differentiating csc(x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not simplifying the equation

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Students may forget to simplify the equation, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result, especially when solving using the product or chain rule. Ensure that each step is written in order. Students might think it is awkward, but it is important to avoid errors in the process.

Mistake 2

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Forgetting the Undefined Points of csc(x)

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They might not remember that csc(x) is undefined at points where sin(x) = 0, such as x = 0, π, 2π, etc. Keep in mind that you should consider the domain of the function that you differentiate. It will help you understand that the function is not continuous at such certain points.

Mistake 3

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Incorrect use of Quotient Rule

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While differentiating functions such as csc(x)/x, students misapply the quotient rule.

 

For example: Incorrect differentiation: d/dx (csc x / x) = -csc(x)cot(x)/x². d/dx (u/v) = (v . u’ - u . v’)/ v² (where u = csc(x) and v = x)

Applying the quotient rule, d/dx (csc x/x) = (x.(-csc(x)cot(x)) - csc(x))/x² To avoid this mistake, write the quotient rule without errors. Always check for errors in the calculation and ensure it is properly simplified.

Mistake 4

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Not writing Constants and Coefficients

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There is a common mistake that students at times forget to multiply the constants placed before csc(x).

 

For example, they incorrectly write d/dx (5 csc(x)) = -csc(x)cot(x).

Students should check the constants in the terms and ensure they are multiplied properly.

 

For e.g., the correct equation is d/dx (5 csc(x)) = -5 csc(x)cot(x).

Mistake 5

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Not Applying the Chain Rule

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Students often forget to use the chain rule. This happens when the derivative of the inner function is not considered.

 

For example: Incorrect: d/dx (csc(2x)) = -csc(2x)cot(2x). To fix this error, students should divide the functions into inner and outer parts. Then, make sure that each function is differentiated.

 

For example, d/dx (csc(2x)) = -2csc(2x)cot(2x).

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Examples Using the Derivative of csc(x)

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Problem 1

Calculate the derivative of (csc(x)·cot(x))

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Here, we have f(x) = csc(x)·cot(x).

Using the product rule, f'(x) = u′v + uv′ In the given equation, u = csc(x) and v = cot(x).

Let’s differentiate each term, u′ = d/dx (csc(x)) = -csc(x)cot(x) v′ = d/dx (cot(x)) = -csc²(x)

Substituting into the given equation, f'(x) = (-csc(x)cot(x))·(cot(x)) + (csc(x))·(-csc²(x))

Let’s simplify terms to get the final answer, f'(x) = -csc(x)cot²(x) - csc³(x)

Thus, the derivative of the specified function is -csc(x)cot²(x) - csc³(x).

Explanation

We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.

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Problem 2

The elevation of a satellite dish is represented by the function y = csc(x), where y represents the height at angle x. If x = π/4 radians, measure the rate of change of the height.

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We have y = csc(x) (height of the satellite dish)...(1)

Now, we will differentiate the equation (1)

Take the derivative csc(x): dy/dx = -csc(x)cot(x)

Given x = π/4 (substitute this into the derivative) dy/dx = -csc(π/4)cot(π/4) csc(π/4) = √2 and cot(π/4) = 1 dy/dx = -√2

Hence, we get the rate of change of the height at x = π/4 as -√2.

Explanation

We find the rate of change of the height at x = π/4 as -√2, which means that at this angle, the height decreases at a rate of √2 units.

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Problem 3

Derive the second derivative of the function y = csc(x).

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The first step is to find the first derivative, dy/dx = -csc(x)cot(x)...(1)

Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-csc(x)cot(x)]

Here we use the product rule, d²y/dx² = -[csc(x)·d/dx(cot(x)) + cot(x)·d/dx(csc(x))] d²y/dx² = -[csc(x)(-csc²(x)) + cot(x)(-csc(x)cot(x))] d²y/dx² = csc³(x) + csc(x)cot²(x)

Therefore, the second derivative of the function y = csc(x) is csc³(x) + csc(x)cot²(x).

Explanation

We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate -csc(x)cot(x). We then substitute the identity and simplify the terms to find the final answer.

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Problem 4

Prove: d/dx (csc²(x)) = -2csc(x)cot(x)csc(x).

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Let’s start using the chain rule:

Consider y = csc²(x) [csc(x)]²

To differentiate, we use the chain rule: dy/dx = 2csc(x)·d/dx[csc(x)]

Since the derivative of csc(x) is -csc(x)cot(x), dy/dx = 2csc(x)(-csc(x)cot(x)) dy/dx = -2csc²(x)cot(x)

Substituting y = csc²(x), d/dx(csc²(x)) = -2csc²(x)cot(x)

Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace csc(x) with its derivative. As a final step, we substitute y = csc²(x) to derive the equation.

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Problem 5

Solve: d/dx (csc(x)/x)

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To differentiate the function, we use the quotient rule: d/dx (csc(x)/x) = (d/dx (csc(x)).x - csc(x).d/dx(x))/x²

We will substitute d/dx (csc(x)) = -csc(x)cot(x) and d/dx (x) = 1 = (x(-csc(x)cot(x)) - csc(x))/x² = (-x csc(x)cot(x) - csc(x))/x² = -(x csc(x)cot(x) + csc(x))/x²

Therefore, d/dx (csc(x)/x) = -(x csc(x)cot(x) + csc(x))/x²

Explanation

In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.

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FAQs on the Derivative of csc(x)

1.Find the derivative of csc(x).

Using the definition of csc(x) as 1/sin(x), the derivative is: d/dx (csc x) = -csc(x)cot(x) (simplified)

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2.Can we use the derivative of csc(x) in real life?

Yes, we can use the derivative of csc(x) in real life to calculate the rate of change of angles, especially in fields such as engineering, physics, and navigation.

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3.Is it possible to take the derivative of csc(x) at the point where x = 0?

No, x = 0 is a point where csc(x) is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).

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4.What rule is used to differentiate csc(x)/x?

We use the quotient rule to differentiate csc(x)/x: d/dx (csc(x)/x) = (x(-csc(x)cot(x)) - csc(x))/x².

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5.Are the derivatives of csc(x) and csc⁻¹(x) the same?

No, they are different. The derivative of csc(x) is -csc(x)cot(x), while the derivative of csc⁻¹(x) is -1/(|x|√(x²-1)).

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Important Glossaries for the Derivative of csc(x)

  • Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.

 

  • Cosecant Function: The cosecant function is one of the primary six trigonometric functions and is written as csc(x).

 

  • Cotangent Function: A trigonometric function that is the reciprocal of the tangent function. It is typically represented as cot(x).

 

  • First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function.

 

  • Asymptote: The line that a graph of a function approaches but never touches or crosses is known as an asymptote.
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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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