Derivative of 1/sin x
2026-02-28 06:07 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-1/sinx

We can derive the derivative of 1/sin 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 1/sin x results in -cot(x)csc(x) using the above-mentioned methods:

By First Principle

The derivative of 1/sin x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of 1/sin x using the first principle, we will consider f(x) = 1/sin 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) = 1/sin x, we write f(x + h) = 1/sin (x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [1/sin(x + h) - 1/sin x] / h = limₕ→₀ [sin x - sin(x + h)] / [h sin x 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 sin(x + h)] = limₕ→₀ [-2 cos(x + h/2) sin(h/2)] / [h sin x sin(x + h)]

Using limit formulas, limₕ→₀ (sin(h/2))/(h/2) = 1. f'(x) = [-2 cos(x)] / [sin²x]

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

Hence, proved.

Using Chain Rule

To prove the differentiation of 1/sin x using the chain rule, We use the formula: 1/sin x = (sin x)⁻¹

Consider f(x) = sin x Using the chain rule: d/dx [f(x)⁻¹] = -[f(x)]⁻² f'(x)

Let’s substitute f(x) = sin x, d/dx (sin x)⁻¹ = -[sin x]⁻² cos x = -cos x/sin²x

Since cot x = cos x/sin x and csc x = 1/sin x, d/dx (1/sin x) = -cot(x)csc(x)

Using Product Rule

We will now prove the derivative of 1/sin x using the product rule.

The step-by-step process is demonstrated below:

Here, we use the formula, 1/sin 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

Here we use the chain rule: v = (sin x)⁻¹ = (sin x)⁻¹ v' = -1. (sin x)⁻² (cos x) v' = -cos x/sin²x Again, use the product rule formula: d/dx (1/sin x) = u'. v + u. v'

Let’s substitute u = 1, u' = 0, v = (sin x)⁻¹, and v' = -cos x/sin²x

When we simplify each term: We get, d/dx (1/sin x) = -cos x/sin²x = -cot(x)csc(x)