Rivalry2

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Original 2026-01-01

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1 We can derive the derivative of 3tanx 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:

2 <ul><li>By First Principle</li>

3 <li>Using Chain Rule</li>

4 <li>Using Product Rule</li>

5 </ul>We will now demonstrate that the differentiation of 3tanx results in 3sec²x using the above-mentioned methods:

6 <h3>By First Principle</h3>

7 The derivative of 3tanx can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.

8 To find the derivative of 3tanx using the first principle, we will consider f(x) = 3tanx.

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

10 Given that f(x) = 3tanx, we write f(x + h) = 3tan (x + h).

11 Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [3tan(x + h) - 3tanx] / h = 3limₕ→₀ [tan(x + h) - tanx] / h = 3limₕ→₀ [ [sin (x + h) / cos (x + h)] - [sin x / cos x] ] / h = 3limₕ→₀ [ [sin (x + h ) cos x - cos (x + h) sin x] / [cos x · cos(x + h)] ]/ h

12 We now use the formula sin A cos B - cos A sin B = sin (A - B). f'(x) = 3limₕ→₀ [ sin (x + h - x) ] / [ h cos x · cos(x + h)] = 3limₕ→₀ [ sin h ] / [ h cos x · cos(x + h)] = 3limₕ→₀ (sin h)/ h · limₕ→₀ 1 / [cos x · cos(x + h)]

13 Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = 3 [ 1 / (cos x · cos(x + 0))] = 3/cos²x

14 As the reciprocal of cosine is secant, we have, f'(x) = 3sec²x. Hence, proved.

15 <h3>Using Chain Rule</h3>

16 To prove the differentiation of 3tanx using the chain rule, We use the formula: 3tanx = 3(sin x/ cos x)

17 Consider f(x) = sin x and g (x)= cos x So we get, 3tanx = 3(f(x)/ g(x))

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

19 Let’s substitute f(x) = sin x and g (x) = cos x in equation (1),

20 d/dx (3tanx) = 3[(cos x) (cos x)- (sin x) (- sin x)]/ (cos x)² = 3(cos²x + sin²x)/ cos²x …(2)

21 Here, we use the formula: (cos²x) + (sin²x) = 1 (Pythagorean identity)

22 Substituting this into (2), d/dx (3tanx) = 3/ (cos x)² Since sec x = 1/cos x, we write: d/dx(3tanx) = 3sec²x

23 <h3>Using Product Rule</h3>

24 We will now prove the derivative of 3tanx using the<a>product</a>rule. The step-by-step process is demonstrated below:

25 Here, we use the formula, 3tanx = 3(sin x/ cos x) = 3(sin x). (cos x)⁻¹

26 Given that, u = 3sin x and v = (cos x)⁻¹

27 Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (3sin x) = 3cos x. (substitute u = 3sin x)

28 Here we use the chain rule: v = (cos x)⁻¹ = (cos x)⁻¹ (substitute v = (cos x)⁻¹) v' = -1. (cos)⁻². d/dx (cos x) v' = sin x/ (cos x)² Again, use the product rule formula: d/dx (3tanx) = u'. v + u. v'

29 Let’s substitute u = 3sin x, u' = 3cos x, v = (cos x)⁻¹, and v' = sin x/ (cos x)²

30 When we simplify each<a>term</a>: We get, d/dx (3tanx) = 3 + 3sin²x / (cos x)² = 3(1 + tan²x) (we use the identity sin²x + cos²x =1)

31 Thus: d/dx (3tanx) = 3 + 3tan²x Since, 3 + 3tan²x = 3sec²x d/dx (3tanx) = 3sec²x.