Last updated on July 21st, 2025
We use the derivative of arccsc(x), which is -1/|x|√(x²-1), as a tool to measure how the arccosecant 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 arccsc(x) in detail.
We now understand the derivative of arccsc(x). It is commonly represented as d/dx (arccsc x) or (arccsc x)', and its value is -1/|x|√(x²-1).
The function arccsc(x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Arccosecant Function: arccsc(x) is the inverse of the cosecant function.
Inverse Function Rule: Rule for differentiating the inverse trigonometric functions.
Absolute Value: The derivative includes an absolute value to ensure positivity of the denominator.
The derivative of arccsc(x) can be denoted as d/dx (arccsc x) or (arccsc x)'.
The formula we use to differentiate arccsc(x) is: d/dx (arccsc x) = -1/|x|√(x²-1)
The formula applies to all x where |x| > 1.
We can derive the derivative of arccsc(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 Implicit Differentiation We will now demonstrate that the differentiation of arccsc(x) results in -1/|x|√(x²-1)
using the above-mentioned methods: By First Principle The derivative of arccsc(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of arccsc(x) using the first principle, we will consider f(x) = arccsc(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) = arccsc(x), we write f(x + h) = arccsc(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [arccsc(x + h) - arccsc(x)] / h = limₕ→₀ [csc⁻¹(x + h) - csc⁻¹(x)] / h This can be complex, so we use implicit differentiation for a simpler approach.
Using Chain Rule To prove the differentiation of arccsc(x) using the chain rule, We use the identity: arccsc(x) = sin⁻¹(1/x) Let y = arccsc(x) implies x = csc(y) = 1/sin(y)
Differentiating both sides with respect to x, we have: 1 = -csc(y)cot(y) dy/dx dy/dx = -1/[csc(y)cot(y)] Since csc(y) = x, and cot(y) = √(x²-1), we have: dy/dx = -1/[x√(x²-1)]
Using Implicit Differentiation We will now prove the derivative of arccsc(x) using implicit differentiation.
The step-by-step process is demonstrated below: Let y = arccsc(x) implies x = csc(y) Taking derivatives on both sides with respect to x, we have: 1 = -csc(y)cot(y) dy/dx dy/dx = -1/[csc(y)cot(y)] = -1/[x√(x²-1)] Hence, proved.
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 arccsc(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 arccsc(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).
When x is 1 or -1, the derivative is undefined because arccsc(x) has a vertical asymptote there. When x tends to infinity, the derivative of arccsc(x) approaches 0.
Students frequently make mistakes when differentiating arccsc(x).
These mistakes can be resolved by understanding the proper solutions.
Here are a few common mistakes and ways to solve them:
Calculate the derivative of (arccsc(x)·x²)
Here, we have f(x) = arccsc(x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = arccsc(x) and v = x².
Let’s differentiate each term, u′ = d/dx (arccsc(x)) = -1/|x|√(x²-1) v′ = d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (-1/|x|√(x²-1))·(x²) + (arccsc(x))·(2x)
Let’s simplify terms to get the final answer, f'(x) = -x/|x|√(x²-1) + 2x·arccsc(x)
Thus, the derivative of the specified function is -x/|x|√(x²-1) + 2x·arccsc(x).
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.
A new bridge is being built over a river with its height modeled by the function y = arccsc(x), where y represents the height of the bridge at a distance x from the shore. If x = 2 meters, measure the rate of change of the bridge's height.
We have y = arccsc(x) (height of the bridge)...(1) Now, we will differentiate the equation (1)
Take the derivative of arccsc(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√(4-1) = -1/2√3
Hence, the rate of change of the bridge's height at a distance x = 2 is -1/2√3.
We find the rate of change of the bridge's height at x = 2 as -1/2√3, which indicates that the height decreases as the distance from the shore increases.
Derive the second derivative of the function y = arccsc(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)]
Here we use the quotient rule and chain rule, d²y/dx² = d/dx [-1]·[1/|x|√(x²-1)] + [-1]·d/dx [1/|x|√(x²-1)]
Simplifying, we get a complex expression for the second derivative.
Therefore, the second derivative of the function y = arccsc(x) involves further application of the chain rule and is more complex.
We use the step-by-step process, starting with the first derivative. Using the quotient rule, we differentiate -1/|x|√(x²-1). We then continue to apply differentiation rules to find the second derivative.
Prove: d/dx (arccsc(x²)) = -2x/|x²|√(x⁴-1).
Let’s start using the chain rule: Consider y = arccsc(x²) To differentiate, we use the chain rule: dy/dx = d/dx [arccsc(u)] where u = x² dy/dx = d/dx [arccsc(u)]·du/dx Since the derivative of arccsc(u) is -1/|u|√(u²-1) and du/dx = 2x, dy/dx = -1/|x²|√(x⁴-1)·2x Therefore, d/dx (arccsc(x²)) = -2x/|x²|√(x⁴-1).
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replaced the inner function with its derivative.
As a final step, we substituted y = arccsc(x²) to derive the equation.
Solve: d/dx (arccsc(x)/x)
To differentiate the function, we use the quotient rule: d/dx (arccsc(x)/x) = (d/dx (arccsc(x))·x - arccsc(x)·d/dx(x))/x² We substitute d/dx (arccsc(x)) = -1/|x|√(x²-1) and d/dx(x) = 1 = (-1/|x|√(x²-1)·x - arccsc(x)·1)/x² = (-1/√(x²-1) - arccsc(x))/x² Therefore, d/dx (arccsc(x)/x) = (-1/√(x²-1) - arccsc(x))/x²
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.
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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