Derivative of 5tanx
2026-02-28 08:26 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-5tanx

We can derive the derivative of 5tanx 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 5tanx results in 5sec²x using the above-mentioned methods:

By First Principle

The derivative of 5tanx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 5tanx using the first principle, we will consider f(x) = 5tanx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 5tanx, we write f(x + h) = 5tan(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [5tan(x + h) - 5tanx] / h = 5 limₕ→₀ [tan(x + h) - tanx] / h = 5 limₕ→₀ [[sin(x + h) / cos(x + h)] - [sinx / cosx]] / h = 5 limₕ→₀ [[sin(x + h)cosx - cos(x + h)sinx] / [cosx · cos(x + h)]] / h We now use the formula sinAcosB - cosAsinB = sin(A-B). f'(x) = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ (sinh)/h · limₕ→₀ 1 / [cosx · cos(x + h)] Using limit formulas, limₕ→₀ (sinh)/h = 1. f'(x) = 5 [1 / (cosx · cos(x + 0))] = 5/cos²x As the reciprocal of cosine is secant, we have, f'(x) = 5sec²x. Hence, proved.

Using Chain Rule

To prove the differentiation of 5tanx using the chain rule, We use the formula: Tanx = sinx/cosx Consider f(x) = sinx and g(x) = cosx So we get, tanx = 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) = sinx and g(x) = cosx in equation (1), d/dx (tanx) = [(cosx)(cosx) - (sinx)(-sinx)] / (cosx)² (cos²x + sin²x) / cos²x …(2) Here, we use the formula: (cos²x) + (sin²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tanx) = 1 / (cosx)² Since secx = 1/cosx, we write: d/dx(tanx) = sec²x Thus, d/dx (5tanx) = 5sec²x

Using Product Rule

We will now prove the derivative of 5tanx using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, 5tanx = 5(sinx/cosx) 5tanx = 5(sinx)(cosx)⁻¹ Given that, u = sinx and v = (cosx)⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (sinx) = cosx. (substitute u = sinx) Here we use the chain rule: v = (cosx)⁻¹ = (cosx)⁻¹ (substitute v = (cosx)⁻¹) v' = -1(cos)⁻² d/dx (cosx) v' = sinx / (cosx)² Again, use the product rule formula: d/dx (5tanx) = 5[u'.v + u.v'] Let’s substitute u = sinx, u' = cosx, v = (cosx)⁻¹, and v' = sinx / (cosx)² When we simplify each term: We get, d/dx (5tanx) = 5(1 + sin²x / (cosx)²) sin²x / (cosx)² = tan²x (we use the identity sin²x + cos²x = 1) Thus: d/dx (5tanx) = 5(1 + tan²x) Since, 1 + tan²x = sec²x d/dx (5tanx) = 5sec²x.