Rivalry2

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Original 2026-01-01

Modified 2026-02-28

1 We can derive the derivative of csc(x) using proofs.

2 To show this, we will use the trigonometric identities along with the rules of differentiation.

3 There are several methods we use to prove this, such as:

4 By First Principle

5 Using Chain Rule

6 Using Product Rule

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

8 By First Principle

9 The derivative of csc(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.

10 To find the derivative of csc(x) using the first principle, we will consider f(x) = csc(x).

11 Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

12 Given that f(x) = csc(x), we write f(x + h) = csc(x + h).

13 Substituting these into<a>equation</a>(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))]

14 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)]

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

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

17 Hence, proved.

18 Using Chain Rule

19 To prove the differentiation of csc(x) using the chain rule,

20 We use the formula: csc(x) = 1/sin(x) Consider f(x) = 1 and g(x) = sin(x)

21 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)

22 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)

23 Using Product Rule

24 We will now prove the derivative of csc(x) using the<a>product</a>rule.

25 The step-by-step process is demonstrated below:

26 Here, we use the formula, csc(x) = 1/sin(x) csc(x) = (sin(x))⁻¹

27 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)

28 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)

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

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

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