Derivative of csc(x)
2026-02-28 06:15 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-csc(x)

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:

By First Principle

Using Chain Rule

Using Product Rule

We will now demonstrate that the differentiation of csc(x) results in -csc(x)cot(x) using the above-mentioned methods:

By First Principle

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)) / (h sin(x) sin(x + h))]

Using the identity 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) sin(x + h)]

Applying limit formulas, limₕ→₀ sin(h/2)/(h/2) = 1, f'(x) = -cos(x) / sin²(x)

As csc(x) = 1/sin(x) and cot(x) = cos(x)/sin(x), we have, f'(x) = -csc(x)cot(x).

Hence, proved.

Using Chain Rule

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)] / sin²(x) = -cos(x) / sin²(x) Since csc(x) = 1/sin(x) and cot(x) = cos(x)/sin(x), we can write: d/dx(csc x) = -csc(x)cot(x)

Using Product Rule

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 = 1 and v = (sin(x))⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (1) = 0 (substitute u = 1)

Here we use the chain rule: v = (sin(x))⁻¹ = (sin(x))⁻¹ (substitute v = (sin(x))⁻¹) v' = -1(sin(x))⁻² · cos(x) v' = -cos(x)/sin²(x)

Again, use the product rule formula: d/dx (csc(x)) = u'.v + u.v'

Let’s substitute u = 1, u' = 0, v = (sin(x))⁻¹, and v' = -cos(x)/sin²(x)

When we simplify each term: We get, d/dx (csc(x)) = -cos(x)/sin²(x) = -csc(x)cot(x)