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

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)] 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)] = limₕ→₀ 2 cos(x + h/2) · (sin(h/2)/(h/2)) · (1/sin x sin(x + h)) Using limit formulas, limₕ→₀ (sin(h/2)/(h/2)) = 1. f'(x) = 2 cos(x + 0) [1/sin x sin x] = 2 cos x / sin² x As the reciprocal of sine is cosecant, 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] / (sin x)² = cos x / sin² x …(2) Here, we use the formula: cos x / sin x = cot x Substituting this into (2), 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)⁻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·d/dx (sin 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) = 0 + -1·(-cos x / sin² x) = cos x / sin² x Since csc x = 1/sin x and cot x = cos x / sin x, we write: d/dx(-csc x) = csc(x)cot(x).