Derivative of -cot x
2026-02-28 10:27 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-minus-cotx

We can derive the derivative of -cot 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 -cot x results in csc²x using the above-mentioned methods:

By First Principle

The derivative of -cot x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

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

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

We now use the formula cos A sin B - sin A cos B = -sin (A - B). f'(x) = limₕ→₀ [- sin (x + h - x) ] / [ h sin x · sin(x + h)] = limₕ→₀ [- sin h ] / [ h sin x · sin(x + h)] = limₕ→₀ -(sin h)/ h · limₕ→₀ 1 / [sin x · sin(x + h)]

Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = -1 [ 1 / (sin x · sin(x + 0))] = 1/sin² x As the reciprocal of sine is cosecant, we have, f'(x) = csc² x.

Hence, proved.

Using Chain Rule

To prove the differentiation of -cot x using the chain rule, We use the formula: -cot x = -cos x/ sin x Consider f(x) = -cos x and g (x)= sin x So we get, -cot 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) = -cos x and g (x) = sin x in equation (1), d/ dx (-cot x) = [(-sin x) (sin x)- (-cos x) (cos x)]/ (sin x)² (-sin² x + cos² x)/ sin² x …(2)

Here, we use the formula: (cos² x) - (sin² x) = cos(2x) (Trigonometric identity) Substituting this into (2), d/dx (-cot x) = 1/ (sin x)² Since csc x = 1/sin x, we write: d/dx(-cot x) = csc² x

Using Product Rule

We will now prove the derivative of -cot x using the product rule.

The step-by-step process is demonstrated below: Here, we use the formula, -cot x = -cos x/ sin x -cot x = (-cos x). (sin x)⁻¹ Given that, u = -cos x and v = (sin x)⁻¹

Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (-cos x) = sin x. (substitute u = -cos x) Here we use the chain rule: v = (sin x)⁻¹ = (sin x)⁻¹ (substitute v = (sin x)⁻¹) v' = -1. (sin)⁻². d/dx (sin x) v' = -cos x/ (sin x)²

Again, use the product rule formula: d/dx (-cot x) = u'. v + u. V' Let’s substitute u = -cos x, u' = sin x, v = (sin x)⁻¹, and v' = -cos x/ (sin x)²

When we simplify each term: We get, d/dx (-cot x) = 1 + cos²x / (sin x)² -cos² x/ (sin x)² = -cot² x (we use the identity cos² x - sin² x =cos(2x)) Thus: d/dx (-cot x) = 1 + cot² x Since, 1 + cot² x = csc² x d/dx (-cot x) = csc² x.