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103 LearnersLast updated on September 15, 2025
Derivative of csc inverse
We use the derivative of csc⁻¹(x), which is -1/(|x|√(x²-1)), as a measuring tool for how the cosecant inverse 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⁻¹(x) in detail.
What is the Derivative of csc inverse?
We now understand the derivative of csc⁻¹(x). It is commonly represented as d/dx (csc⁻¹(x)) or (csc⁻¹(x))', and its value is -1/(|x|√(x²-1)). The function csc⁻¹(x) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Cosecant Inverse Function: (csc⁻¹(x)).
Chain Rule: Rule for differentiating complex functions (useful for csc⁻¹(x)).
Absolute Value Function: Represents the absolute value operation, important in the derivative.
Derivative of csc inverse Formula
The derivative of csc⁻¹(x) can be denoted as d/dx (csc⁻¹(x)) or (csc⁻¹(x))'.
The formula we use to differentiate csc⁻¹(x) is: d/dx (csc⁻¹(x)) = -1/(|x|√(x²-1))
The formula applies to all x where |x| > 1.
Proofs of the Derivative of csc inverse
We can derive the derivative of csc⁻¹(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 Chain Rule
To prove the differentiation of csc⁻¹(x) using the chain rule, we start with the identity: y = csc⁻¹(x) x = csc(y) Differentiating both sides with respect to x gives: 1 = -csc(y)cot(y)dy/dx
Thus, dy/dx = -1/(csc(y)cot(y)) Since x = csc(y), cot²(y) = csc²(y) - 1 = x² - 1
This implies: dy/dx = -1/(√(x²-1)x) Thus, dy/dx = -1/(|x|√(x²-1)).
Higher-Order Derivatives of csc inverse
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⁻¹(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⁻¹(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).
Special Cases:
When x is 1 or -1, the derivative is undefined because csc⁻¹(x) has vertical asymptotes there.
When x is √2, the derivative of csc⁻¹(x) = -1/ (√2√(2-1)) = -1.
Common Mistakes and How to Avoid Them in Derivatives of csc inverse
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
Not simplifying the equation
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 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
Forgetting the Undefined Points of csc inverse
They might not remember that csc⁻¹(x) is undefined at points such as x = 1 and x = -1. 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
Incorrect use of Chain Rule
While differentiating functions such as csc⁻¹(2x), students misapply the chain rule. For example: Incorrect differentiation: d/dx (csc⁻¹(2x)) = -1/(|2x|√((2x)²-1)). To avoid this mistake, write the chain rule without errors. Always check for errors in the calculation and ensure it is properly simplified.
Mistake 4
Not writing Constants and Coefficients
There is a common mistake that students at times forget to multiply the constants before csc⁻¹(x). For example, they incorrectly write d/dx (5 csc⁻¹(x)) = -1/(|x|√(x²-1)). 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⁻¹(x)) = -5/(|x|√(x²-1)).
Mistake 5
Not Applying the Absolute Value
Students often forget to use the absolute value, which is crucial in the derivative of csc⁻¹(x). For example: Incorrect: d/dx (csc⁻¹(x)) = -1/(x√(x²-1)). To fix this error, ensure the absolute value is used for x. For example, d/dx (csc⁻¹(x)) = -1/(|x|√(x²-1)).
Examples Using the Derivative of csc inverse
Problem 1
Calculate the derivative of (3csc⁻¹(x) + x²).
Here, we have f(x) = 3csc⁻¹(x) + x². Using the sum rule, f'(x) = d/dx [3csc⁻¹(x)] + d/dx [x²]
For the first term, we use the derivative of csc⁻¹(x): d/dx [3csc⁻¹(x)] = -3/(|x|√(x²-1))
For the second term: d/dx [x²] = 2x
Thus, f'(x) = -3/(|x|√(x²-1)) + 2x.
Thus, the derivative of the specified function is -3/(|x|√(x²-1)) + 2x.
Explanation
We find the derivative of the given function by separating it into two parts. The first step is finding its derivative and then combining them using the sum rule to get the final result.
Problem 2
PQR Solutions is analyzing a curve represented by the function y = csc⁻¹(x) for a particular material's stress-strain relationship. At what rate does the stress change when the strain is x = 2?
We have y = csc⁻¹(x) (stress of the material)...(1)
Now, we will differentiate the equation (1) Take the derivative csc⁻¹(x): dy/dx = -1/(|x|√(x²-1))
Given x = 2, substitute this into the derivative: dy/dx = -1/(|2|√(2²-1)) dy/dx = -1/(2√3)
Hence, the rate of change of stress at a strain of x = 2 is -1/(2√3).
Explanation
We find the rate of change of stress at x = 2 as -1/(2√3), which indicates how quickly the stress changes for a given strain value.
Problem 3
Derive the second derivative of the function y = csc⁻¹(x).
The first step is to find the first derivative, dy/dx = -1/(|x|√(x²-1))...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(|x|√(x²-1))]
Using the quotient and chain rules, d²y/dx² = [d/dx (-1) × |x|√(x²-1) - (-1) × d/dx (|x|√(x²-1))] / (|x|²(x²-1))
This requires a detailed chain rule application, resulting in a complex expression.
Therefore, the second derivative of the function y = csc⁻¹(x) involves a detailed calculation beyond basic steps.
Explanation
We use the step-by-step process, where we start with the first derivative. Using the quotient and chain rules, we differentiate -1/(|x|√(x²-1)). The second derivative involves complex expressions that require careful application of differentiation rules.
Problem 4
Prove: d/dx ((csc⁻¹(x))²) = 2csc⁻¹(x) × (-1/(|x|√(x²-1))).
Let’s start using the chain rule: Consider y = (csc⁻¹(x))²
To differentiate, we use the chain rule: dy/dx = 2csc⁻¹(x) × d/dx [csc⁻¹(x)]
Since the derivative of csc⁻¹(x) is -1/(|x|√(x²-1)), dy/dx = 2csc⁻¹(x) × (-1/(|x|√(x²-1)))
Thus, d/dx ((csc⁻¹(x))²) = 2csc⁻¹(x) × (-1/(|x|√(x²-1))).
Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replaced csc⁻¹(x) with its derivative. As a final step, we simplified to derive 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) = [d/dx (csc⁻¹(x)) × x - csc⁻¹(x) × d/dx(x)] / x²
We will substitute d/dx (csc⁻¹(x)) = -1/(|x|√(x²-1)) and d/dx (x) = 1 = [-1/(|x|√(x²-1)) × x - csc⁻¹(x)] / 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. As a final step, we simplify the equation to obtain the final result.
FAQs on the Derivative of csc inverse
1.Find the derivative of csc⁻¹(x).
2.Can we use the derivative of csc inverse in real life?
3.Is it possible to take the derivative of csc inverse at the point where x = 1?
4.What rule is used to differentiate csc inverse divided by x?
5.Are the derivatives of csc⁻¹(x) and sin⁻¹(x) the same?
Important Glossaries for the Derivative of csc inverse
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Cosecant Inverse Function: The inverse function of cosecant, denoted as csc⁻¹(x).
- Chain Rule: A rule used in calculus for differentiating the composition of two or more functions.
- Absolute Value: The non-negative value of a number without regard to its sign, denoted as |x|.
- Quotient Rule: A method of finding the derivative of the division of two functions.
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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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