Rivalry2

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Original 2026-01-01

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1 We can derive the derivative of -cot 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 <ul><li>By First Principle</li>

5 <li>Using Chain Rule</li>

6 <li>Using Product Rule</li>

7 </ul>We will now demonstrate that the differentiation of -cot x results in csc²x using the above-mentioned methods:

8 By First Principle

9 The derivative of -cot 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 -cot x using the first principle, we will consider f(x) = -cot x.

11 Its derivative can be expressed as the following limit.

12 f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

13 Given that f(x) = -cot x, we write f(x + h) = -cot (x + h).

14 Substituting these into<a>equation</a>(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

15 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)]

16 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.

17 Hence, proved.

18 Using Chain Rule

19 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)

20 By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1)

21 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)

22 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

23 Using Product Rule

24 We will now prove the derivative of -cot x using the<a>product</a>rule.

25 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

26 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

27 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

28 When we simplify each<a>term</a>: 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.