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Last updated on October 9, 2025

Derivative of csc x²

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We use the derivative of csc(x²) as a measuring tool for how the cosecant function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of csc(x²) in detail.

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What is the Derivative of csc x²?

We now understand the derivative of csc x². It is commonly represented as d/dx (csc x²) or (csc x²)', and its value can be derived using the chain and product rules. The function csc x² 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).

 

Chain Rule: A rule for differentiating composite functions, such as csc(x²).

 

Quotient Rule: Another rule that can be applied when differentiating functions expressed as a ratio.

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Derivative of csc x² Formula

The derivative of csc x² can be denoted as d/dx (csc x²) or (csc x²)'.

 

The formula we use to differentiate csc x² involves using the chain rule: d/dx (csc x²) = -2x csc(x²) cot(x²) The formula applies to all x where sin(x²) ≠ 0.

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Proofs of the Derivative of csc x²

We can derive the derivative of csc x² 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 Chain Rule
     
  • Using Quotient Rule

 

We will now demonstrate that the differentiation of csc x² results in -2x csc(x²) cot(x²) using the above-mentioned methods:

 

Using Chain Rule

 

To prove the differentiation of csc x² using the chain rule, Consider f(x) = csc(x²) = 1/sin(x²). Let g(x) = x², then f(g(x)) = csc(g(x)). The chain rule gives us: d/dx [f(g(x))] = f'(g(x))g'(x). Here, f'(u) = -csc(u) cot(u) and g'(x) = 2x. Thus, d/dx (csc x²) = -csc(x²) cot(x²) * 2x = -2x csc(x²) cot(x²).

 

Using Quotient Rule

 

We can also use the quotient rule to find the derivative: csc(x²) = 1/sin(x²). Let u = 1 and v = sin(x²). The quotient rule states: d/dx (u/v) = (v u' - u v')/v². u' = 0 and v' = cos(x²) * 2x. Thus, d/dx (csc x²) = (sin(x²) * 0 - 1 * cos(x²) * 2x) / (sin(x²))² = -2x cos(x²) / sin²(x²) = -2x csc(x²) cot(x²).

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Higher-Order Derivatives of csc x²

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(x²).

 

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(x²), 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 such that x² = nπ (where n is an integer), the derivative is undefined because csc(x²) has a vertical asymptote there. When x is 0, the derivative of csc x² = -2x csc(x²) cot(x²), which is 0.

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Common Mistakes and How to Avoid Them in Derivatives of csc x²

Students frequently make mistakes when differentiating csc x². 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 x²

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They might not remember that csc x² is undefined at points such as (x² = nπ, where n is an integer). 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(x²)/x, students misapply the quotient rule. For example: Incorrect differentiation: d/dx (csc(x²) / x) = -2x csc(x²) cot(x²) / x². d/dx (u/v) = (v u' - u v')/v² (where u = csc(x²) and v = x) Applying the quotient rule, d/dx (csc(x²)/x) = (x(-2x csc(x²) cot(x²)) - csc(x²))/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 x². For example, they incorrectly write d/dx (5 csc x²) = -2x csc(x²) cot(x²). 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 x²) = 5(-2x csc(x²) cot(x²)).

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(2x²)) = -2x csc(2x²) cot(2x²). 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(2x²)) = -4x csc(2x²) cot(2x²).

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Examples Using the Derivative of csc x²

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Problem 1

Calculate the derivative of (csc(x²)·sec(x)).

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Here, we have f(x) = csc(x²)·sec(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = csc(x²) and v = sec(x). Let’s differentiate each term, u′= d/dx (csc(x²)) = -2x csc(x²) cot(x²) v′= d/dx (sec(x)) = sec(x) tan(x) Substituting into the given equation, f'(x) = (-2x csc(x²) cot(x²)) sec(x) + csc(x²) (sec(x) tan(x)) Let’s simplify terms to get the final answer, f'(x) = -2x csc(x²) cot(x²) sec(x) + csc(x²) sec(x) tan(x) Thus, the derivative of the specified function is -2x csc(x²) cot(x²) sec(x) + csc(x²) sec(x) tan(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

An engineering firm is analyzing the stress on a beam, modeled by y = csc(x²), where y represents stress and x is the length. If x = 1 meter, calculate the rate of change of stress.

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We have y = csc(x²) (stress on the beam)...(1) Now, we will differentiate the equation (1) Take the derivative csc(x²): dy/dx = -2x csc(x²) cot(x²) Given x = 1 (substitute this into the derivative) dy/dx = -2(1) csc(1²) cot(1²) = -2 csc(1) cot(1) Hence, we get the rate of change of stress at x=1 as -2 csc(1) cot(1).

Explanation

We find the rate of change of stress at x=1 using the derivative formula.

This gives insight into how stress changes with respect to length at a specific point.

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Problem 3

Derive the second derivative of the function y = csc(x²).

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The first step is to find the first derivative, dy/dx = -2x csc(x²) cot(x²)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-2x csc(x²) cot(x²)] Apply the product rule, d²y/dx² = -2[csc(x²) cot(x²) + x(d/dx (csc(x²) cot(x²)))] We find d/dx (csc(x²) cot(x²)) using product and chain rules, = d/dx (csc(x²)) cot(x²) + csc(x²) d/dx (cot(x²)) = [-2x csc(x²) cot²(x²) - 2x csc(x²) csc²(x²)] Substitute back, d²y/dx² = -2[csc(x²) cot(x²) - 2x csc(x²) cot²(x²) - 2x csc(x²) csc²(x²)] Therefore, the second derivative of the function y = csc(x²) is a more complex expression involving trigonometric identities.

Explanation

We use the step-by-step process, where we start with the first derivative.

Using the product and chain rules, we differentiate to find the second derivative.

The expression involves multiple trigonometric functions.

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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) 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(x²)/x)

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To differentiate the function, we use the quotient rule: d/dx (csc(x²)/x) = (d/dx (csc(x²)) · x - csc(x²) · d/dx(x))/x² We will substitute d/dx (csc(x²)) = -2x csc(x²) cot(x²) and d/dx (x) = 1 = (-2x csc(x²) cot(x²) x - csc(x²))/x² = (-2x² csc(x²) cot(x²) - csc(x²))/x² Therefore, d/dx (csc(x²)/x) = (-2x² csc(x²) cot(x²) - csc(x²))/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 csc x²

1.Find the derivative of csc x².

Using the chain rule on csc x² gives: d/dx (csc x²) = -2x csc(x²) cot(x²).

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2.Can we use the derivative of csc x² in real life?

Yes, we can use the derivative of csc x² in real life to analyze rates of change in trigonometric models, especially in engineering and physics.

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3.Is it possible to take the derivative of csc x² at points where x² = nπ?

No, x² = nπ is a point where csc x² 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(x²) / x?

We use the quotient rule to differentiate csc(x²)/x, d/dx (csc(x²)/x) = (x(-2x csc(x²) cot(x²)) - csc(x²))/x².

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5.Are the derivatives of csc(x²) and csc²(x) the same?

No, they are different. The derivative of csc(x²) is -2x csc(x²) cot(x²), while the derivative of csc²(x) is -2 csc²(x) cot(x).

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Important Glossaries for the Derivative of csc x²

  • 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 a trigonometric function, represented as csc(x) = 1/sin(x).

 

  • Chain Rule: A rule used to differentiate the composition of functions.

 

  • Quotient Rule: A method to differentiate functions expressed as a quotient of two functions.

 

  • Vertical Asymptote: A line where a function approaches but never touches or crosses, indicating points of discontinuity. ```
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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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