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Last updated on September 27, 2025
We use the derivative of csc(x), which is -csc(x)cot(x), 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(x) in detail.
We now understand the derivative of csc x. It is commonly represented as d/dx (csc x) or (csc x)', and its value is -csc(x)cot(x). The function csc x 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)).
Quotient Rule: Rule for differentiating csc(x) (since it consists of 1/sin(x)).
Cotangent Function: cot(x) = cos(x)/sin(x).
The derivative of csc x can be denoted as d/dx (csc x) or (csc x)'. The formula we use to differentiate csc x is: d/dx (csc x) = -csc(x)cot(x) (or) (csc x)' = -csc(x)cot(x)
The formula applies to all x where sin(x) ≠ 0
We can derive the derivative of csc 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:
We will now demonstrate that the differentiation of csc x results in -csc(x)cot(x) using the above-mentioned methods:
The derivative of csc x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of csc x using the first principle, we will consider f(x) = csc 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) = csc x, we write f(x + h) = csc (x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [csc(x + h) - csc x] / h = limₕ→₀ [ [1/sin (x + h)] - [1/sin x] ] / h = limₕ→₀ [ [sin x - sin (x + h)] / [sin(x + h) · sin x] ] / h We now use the formula sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2). f'(x) = limₕ→₀ [ -2 cos((2x + h)/2) sin(h/2) ] / [ h sin(x + h) · sin x] = limₕ→₀ [ -2 cos(x + h/2) sin(h/2) / h ] · [ 1 / sin(x + h) · sin x ] Using limit formulas, limₕ→₀ (sin(h/2)/(h/2)) = 1. f'(x) = -2 cos(x) / [sin²x] As cosecant is the reciprocal of sine, we have, f'(x) = -csc(x)cot(x). Hence, proved.
To prove the differentiation of csc x using the chain rule, We use the formula: Csc x = 1/sin x Consider f(x) = 1 and g(x)= sin x So we get, csc x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1) Let’s substitute f(x) = 1 and g(x) = sin x in equation (1), d/dx (csc x) = [(0) (sin x) - (1) (cos x)] / (sin x)² = -cos x / sin² x Here, we use the formula: cos x / sin x = cot x (cotangent identity) Substituting this into the equation, d/dx (csc x) = -csc(x)cot(x)
We will now prove the derivative of csc x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, Csc x = 1/sin x csc x = (sin x)⁻¹ Given that, u = (sin x)⁻¹ Using the product rule formula: d/dx [u] = -1(sin x)⁻² (cos x) = -cos x / sin² x Here we use the chain rule: v = cos x / sin x = cot x Substituting u = (sin x)⁻¹ and using the product rule, we find: d/dx (csc x) = -csc(x)cot(x).
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(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(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 0, the derivative is undefined because csc(x) has a vertical asymptote there.
When x is π/2, the derivative of csc x = -csc(π/2)cot(π/2), which is 0.
Students frequently make mistakes when differentiating csc 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 (csc x·cot x)
Here, we have f(x) = csc x·cot x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = csc x and v = cot x. Let’s differentiate each term, u′= d/dx (csc x) = -csc(x)cot(x) v′= d/dx (cot x) = -csc²x Substituting into the given equation, f'(x) = (-csc(x)cot(x))·(cot x) + (csc x)·(-csc²x) Let’s simplify terms to get the final answer, f'(x) = -csc(x)cot²x - csc³x Thus, the derivative of the specified function is -csc(x)cot²x - csc³x.
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.
XYZ Finance Company analyzes the risk of investments using the function y = csc(x), where y represents the risk level at a given time x. If x = π/3 years, measure the risk level change.
We have y = csc(x) (risk level function)...(1) Now, we will differentiate the equation (1) Take the derivative csc(x): dy/dx = -csc(x)cot(x) Given x = π/3 (substitute this into the derivative) dy/dx = -csc(π/3)cot(π/3) = -2√3/3 (since csc(π/3) = 2/√3 and cot(π/3) = 1/√3) Hence, we get the risk level change at x= π/3 as -2√3/3.
We find the risk level change at x= π/3 as -2√3/3, which means that at a given time, the risk level would decrease at this rate.
Derive the second derivative of the function y = csc(x).
The first step is to find the first derivative, dy/dx = -csc(x)cot(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-csc(x)cot(x)] Here we use the product rule, d²y/dx² = -[csc(x) d/dx(cot(x)) + cot(x) d/dx(csc(x))] = -[-csc(x)csc²(x) - cot²(x)csc(x)] = csc(x)(csc²(x) + cot²(x)) Therefore, the second derivative of the function y = csc(x) is csc(x)(csc²(x) + cot²(x)).
We use the step-by-step process, where we start with the first derivative.
Using the product rule, we differentiate -csc(x)cot(x).
We then substitute the identities and simplify the terms to find the final answer.
Prove: d/dx (csc²(x)) = -2csc(x)cot(x)csc(x).
Let’s start using the chain rule: Consider y = csc²(x) = [csc(x)]² To differentiate, we use the chain rule: dy/dx = 2csc(x) d/dx [csc(x)] Since the derivative of csc(x) is -csc(x)cot(x), dy/dx = 2csc(x) [-csc(x)cot(x)] Substituting y = csc²(x), d/dx (csc²(x)) = -2csc(x)cot(x)csc(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 x/x)
To differentiate the function, we use the quotient rule: d/dx (csc x/x) = (d/dx (csc x)·x - csc x·d/dx(x))/x² We will substitute d/dx (csc x) = -csc(x)cot(x) and d/dx (x) = 1 = (-csc(x)cot(x)·x - csc x·1) / x² = (-x csc(x)cot(x) - csc x)/x² = -x csc(x)cot(x) - csc x / x² Therefore, d/dx (csc x/x) = -x csc(x)cot(x) - csc x / 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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