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Last updated on July 15th, 2025

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Derivative of Cosecant

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We use the derivative of csc(x), which is -csc(x)cot(x), as a measuring tool for how the cosecant function changes in response to a slight change in x. Derivatives help us calculate various rates of change in real-life situations. We will now talk about the derivative of csc(x) in detail.

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What is the Derivative of Cosecant?

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

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Proofs of the Derivative of Cosecant

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)] We now use the formula 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)] = limₕ→₀ [-2 cos((2x + h)/2) sin(h/2)] / [h sin x sin(x + h)] = limₕ→₀ (-2/h) cos((2x + h)/2) sin(h/2) / [sin x sin(x + h)] Using limit formulas, limₕ→₀ sin(h/2)/(h/2) = 1. f'(x) = -cos(x)/[sin(x) sin x] = -cos(x)/sin²x As the reciprocal of sine is cosecant and cosine over sine is cotangent, 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)]/ (sin x)² = - cos x / sin² x …(2) Here, we use the formula: cos x / sin² x = - csc x cot x Substituting this into (2), 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)⁻² d/dx (sin 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) = 0 + (- cos x / sin² x) = - csc x cot x

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Higher-Order Derivatives of Cosecant

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

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Common Mistakes and How to Avoid Them in Derivatives of Cosecant

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 Cosecant

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They might not remember that csc x is undefined at the points such as (x = 0, π,...). 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)) = -2 csc(2x)cot(2x).

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Examples Using the Derivative of Cosecant

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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 + csc² 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

AXB International School sponsored the construction of a bridge. The height is represented by the function y = csc(x) where y represents the elevation of the bridge at a distance x. If x = π/6 meters, measure the slope of the bridge.

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We have y = csc(x) (height of the bridge)...(1) Now, we will differentiate the equation (1) Take the derivative csc(x): dy/dx = -csc(x)cot(x) We know that cot(x) = cos(x)/sin(x) Given x = π/6 (substitute this into the derivative) dy/dx = -csc(π/6)cot(π/6) = -2√3/3 Hence, we get the slope of the bridge at a distance x = π/6 as -2√3/3.

Explanation

We find the slope of the bridge at x = π/6 as -2√3/3, which means that at a given point, the height of the bridge decreases at a rate of 2√3/3 times the horizontal distance.

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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))] = -[csc(x)(-csc²(x)) + cot(x)(-csc(x)cot(x))] = csc³(x) + csc(x)cot²(x) Therefore, the second derivative of the function y = csc(x) is csc(x)(csc²(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 identities and simplify the terms to find the final answer.

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Problem 4

Prove: d/dx (csc²(x)) = -2 csc²(x)cot(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 = 2 csc(x) d/dx[csc(x)] Since the derivative of csc(x) is -csc(x)cot(x), dy/dx = 2 csc(x) (-csc(x)cot(x)) = -2 csc²(x)cot(x) Substituting y = csc²(x), d/dx (csc²(x)) = -2 csc²(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 = (-csc x cot x. x - csc x. 1) / 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 product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.

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FAQs on the Derivative of Cosecant

1.Find the derivative of csc x.

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

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

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

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

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Important Glossaries for the Derivative of Cosecant

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 function goes near a line without intersecting or crossing it. This line 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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