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105 LearnersLast updated on October 8, 2025
Derivative of csc⁻¹
We use the derivative of csc⁻¹(x), which is -1/(|x|√(x²-1)), as a tool for understanding how the inverse cosecant function changes with respect to x. Derivatives are useful in various real-life applications, such as physics and engineering. We will now delve into the derivative of csc⁻¹(x) in detail.
What is the Derivative of csc⁻¹?
The derivative of csc⁻¹(x) is commonly represented as d/dx (csc⁻¹(x)) or (csc⁻¹(x))', and its value is -1/(|x|√(x²-1)). This function has a well-defined derivative, indicating it is differentiable within its domain.
Key concepts are mentioned below:
Inverse Cosecant Function: csc⁻¹(x) is the inverse of the cosecant function.
Derivative Formula: The derivative of csc⁻¹(x) is calculated using its definition and properties.
Absolute Value: The derivative includes an absolute value, ensuring the expression is valid for negative x.
Derivative of csc⁻¹ Formula
The derivative of csc⁻¹(x) can be denoted as d/dx (csc⁻¹(x)) or (csc⁻¹(x))'.
The formula used to differentiate csc⁻¹(x) is: d/dx (csc⁻¹(x)) = -1/(|x|√(x²-1))
This formula applies to all x where |x| > 1.
Proofs of the Derivative of csc⁻¹
We can derive the derivative of csc⁻¹(x) using different methods. We will use trigonometric identities and differentiation rules to prove this.
There are several methods to prove this, such as:
By Trigonometric Identity
Express csc⁻¹(x) in terms of other trigonometric functions and use known derivatives. If y = csc⁻¹(x), then x = csc(y) and dy/dx = -1/(|x|√(x²-1)). By Implicit Differentiation Start with x = csc(y) and differentiate both sides implicitly with respect to x. Differentiate x = 1/sin(y), obtaining dx/dy = -csc(y)cot(y). Then, solve for dy/dx = -1/(x√(x²-1)).
By Chain Rule
Use the relationship with inverse trigonometric functions: y = csc⁻¹(x) implies x = csc(y). Differentiate implicitly and apply chain rule to get dy/dx = -1/(|x|√(x²-1)).
Higher-Order Derivatives of csc⁻¹
Higher-order derivatives involve differentiating a function multiple times. For csc⁻¹(x), the process can be more complex. To understand them better, consider physical motion where the position changes (first derivative) and the rate of change of position (second derivative) also varies.
For the first derivative, we write f′(x) indicating the rate of change. The second derivative, denoted as f′′(x), represents the change in the rate of change. Similarly, the third derivative, f′′′(x), is derived from the second derivative.
For the nth derivative of csc⁻¹(x), we generally use fⁿ(x) to represent the nth derivative, indicating higher-order changes.
Special Cases:
When x = 1 or x = -1, the derivative is undefined because the expression involves division by zero at these points. When x = √2, the derivative of csc⁻¹(x) is -1/(√2√(2-1)).
Common Mistakes and How to Avoid Them in Derivatives of csc⁻¹
Students frequently make mistakes when differentiating csc⁻¹(x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Mistake 1
Ignoring the Absolute Value
Students may forget to include the absolute value in the derivative, leading to incorrect results.
Ensure that the absolute value is present to handle negative x correctly.
Mistake 2
Misidentifying the Domain
Students might not remember that csc⁻¹(x) is only defined for |x| > 1.
Keeping this in mind helps avoid errors related to the undefined domain of the function.
Mistake 3
Incorrect Use of Trigonometric Identities
While working with csc⁻¹(x), students may misapply trigonometric identities. For example, incorrectly differentiating csc⁻¹(x) using incorrect identities can lead to errors.
To avoid this, ensure that all identities are applied correctly and verify each step.
Mistake 4
Neglecting to Simplify Expressions
Students often forget to simplify expressions after differentiating, resulting in complex and incorrect answers.
Ensure that all expressions are simplified to their simplest form for clarity and correctness.
Mistake 5
Forgetting the Chain Rule
Students may overlook the chain rule when differentiating composite functions involving csc⁻¹(x).
Ensure that the chain rule is applied correctly by identifying inner and outer functions and differentiating each.
Examples Using the Derivative of csc⁻¹
Problem 1
Calculate the derivative of (csc⁻¹(x) + 3x).
Here, we have f(x) = csc⁻¹(x) + 3x. Differentiate each term separately: f'(x) = d/dx (csc⁻¹(x)) + d/dx (3x) Using the derivative formula for csc⁻¹(x), f'(x) = -1/(|x|√(x²-1)) + 3. Thus, the derivative of the specified function is -1/(|x|√(x²-1)) + 3.
Explanation
We find the derivative of the given function by differentiating each term separately and applying the derivative formula for csc⁻¹(x).
The result is obtained by combining these derivatives.
Problem 2
A satellite dish is positioned at an angle θ such that θ = csc⁻¹(x), where x is the distance from the base to the top of the dish. If x = 2 meters, calculate the change in angle with respect to x.
We have θ = csc⁻¹(x) (angle of the satellite dish)...(1) Differentiate the equation (1) with respect to x: dθ/dx = -1/(|x|√(x²-1)) Given x = 2, substitute into the derivative: dθ/dx = -1/(|2|√(2²-1)) dθ/dx = -1/(2√3). Hence, the change in angle with respect to x when x = 2 is -1/(2√3).
Explanation
We find the rate of change of the angle θ with respect to x by differentiating the given function and substituting x = 2 into the derivative formula.
The result indicates how the angle changes as the distance varies.
Problem 3
Derive the second derivative of the function y = csc⁻¹(x).
First, find the first derivative: dy/dx = -1/(|x|√(x²-1))...(1) Now differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(|x|√(x²-1))] Applying the quotient rule, d²y/dx² = [x² - 1 + x²]/(|x|³(x²-1)^(3/2)) Therefore, the second derivative of the function y = csc⁻¹(x) is [2x² - 1]/(|x|³(x²-1)^(3/2)).
Explanation
We use the step-by-step process, starting with the first derivative of csc⁻¹(x).
Using the quotient rule, we differentiate again and simplify the terms to find the second derivative.
Problem 4
Prove: d/dx (x·csc⁻¹(x)) = csc⁻¹(x) - x/(|x|√(x²-1)).
Let's start using the product rule: Consider y = x·csc⁻¹(x). To differentiate: dy/dx = d/dx (x)·csc⁻¹(x) + x·d/dx (csc⁻¹(x)) Using the derivative formula for csc⁻¹(x): dy/dx = 1·csc⁻¹(x) + x·(-1/(|x|√(x²-1))) dy/dx = csc⁻¹(x) - x/(|x|√(x²-1)). Hence proved.
Explanation
In this step-by-step process, we used the product rule to differentiate the equation, substituted the derivative of csc⁻¹(x), and simplified the expression to prove the equation.
Problem 5
Solve: d/dx (csc⁻¹(x)/x)
To differentiate the function, we use the quotient rule: d/dx (csc⁻¹(x)/x) = (x·d/dx (csc⁻¹(x)) - csc⁻¹(x)·d/dx(x))/x² Substitute d/dx (csc⁻¹(x)) = -1/(|x|√(x²-1)) and d/dx (x) = 1: = (x·(-1/(|x|√(x²-1))) - csc⁻¹(x)·1)/x² = (-1/√(x²-1) - csc⁻¹(x))/x². Therefore, d/dx (csc⁻¹(x)/x) = (-1/√(x²-1) - csc⁻¹(x))/x².
Explanation
In this process, we differentiate the given function using the quotient rule and substitute the appropriate derivatives.
We then simplify the equation to obtain the final result.
FAQs on the Derivative of csc⁻¹
1.Find the derivative of csc⁻¹(x).
2.Can we use the derivative of csc⁻¹(x) in real life?
3.Is it possible to take the derivative of csc⁻¹(x) when x = 1?
4.What rule is used to differentiate csc⁻¹(x)/x?
5.Are the derivatives of csc⁻¹(x) and sec⁻¹(x) the same?
Important Glossaries for the Derivative of csc⁻¹
- Derivative: The derivative of a function measures how the function changes as its input changes.
- Inverse Cosecant Function: The inverse of the cosecant function, denoted as csc⁻¹(x).
- Quotient Rule: A method for differentiating functions that are ratios of two differentiable functions.
- Absolute Value: A mathematical function that returns the non-negative value of a number, often used in derivative formulas.
- Higher-Order Derivatives: Derivatives taken multiple times, providing insights into the behavior of functions beyond the first derivative.
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Jaskaran Singh Saluja
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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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