Derivative of csc²(x)
2026-02-28 23:52 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-csc%5E2%28x%29

We can derive the derivative of csc²(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 csc²(x) results in -2csc²(x)cot(x) using the above-mentioned methods:

By First Principle

The derivative of csc²(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of csc²(x) using the first principle, we will consider f(x) = csc²(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) = csc²(x), we write f(x + h) = csc²(x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [csc²(x + h) - csc²(x)] / h = limₕ→₀ [1/sin²(x + h) - 1/sin²(x)] / h = limₕ→₀ [[sin²(x) - sin²(x + h)] / [sin²(x)sin²(x + h)]] / h We now use the formula sin²(A) - sin²(B) = (A - B)(A + B). f'(x) = limₕ→₀ [(sin(x + h) - sin(x))(sin(x + h) + sin(x))] / [h sin²(x)sin²(x + h)] = limₕ→₀ [(sin h) cos((x + x + h)/2) 2cos((x - x - h)/2)] / [h sin²(x)sin²(x + h)] = limₕ→₀ (sin h)/ h · limₕ→₀ [cos((x + x + h)/2) · 2cos((x - x - h)/2)] / [sin²(x)sin²(x + h)] Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = -2csc²(x)cot(x) Hence, proved.

Using Chain Rule

To prove the differentiation of csc²(x) using the chain rule, We use the formula: csc²(x) = (csc(x))² Let u = csc(x), so we have u² By chain rule: d/dx (u²) = 2u u' Let’s substitute u = csc(x) and u' = -csc(x)cot(x), d/dx (csc²(x)) = 2(csc(x))(-csc(x)cot(x)) = -2csc²(x)cot(x)

Using Product Rule

We will now prove the derivative of csc²(x) using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, csc²(x) = (csc(x))(csc(x)) Given that, u = csc(x) and v = csc(x) Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (csc(x)) = -csc(x)cot(x) (substitute u = csc(x)) v' = d/dx (csc(x)) = -csc(x)cot(x) (substitute v = csc(x))

Again, use the product rule formula: d/dx (csc²(x)) = u'.v + u.v' Let’s substitute u = csc(x), u' = -csc(x)cot(x), v = csc(x), and v' = -csc(x)cot(x) When we simplify each term: We get, d/dx (csc²(x)) = -csc(x)cot(x)csc(x) + csc(x)(-csc(x)cot(x)) = -2csc²(x)cot(x) Thus: d/dx (csc²(x)) = -2csc²(x)cot(x).