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Last updated on October 9, 2025
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.
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.
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.
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:
We will now demonstrate that the differentiation of csc x² results in -2x csc(x²) cot(x²) using the above-mentioned methods:
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).
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).
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).
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.
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:
Calculate the derivative of (csc(x²)·sec(x)).
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).
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.
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.
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).
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.
Derive the second derivative of the function y = csc(x²).
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.
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.
Prove: d/dx (csc²(x)) = -2 csc²(x) cot(x).
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.
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.
Solve: d/dx (csc(x²)/x)
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.
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.
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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