Derivative of Tan(2x)
2026-02-28 06:07 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-tan(2x)

We can derive the derivative of tan(2x) 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 tan(2x) results in 2sec²(2x) using the above-mentioned methods:

By First Principle

The derivative of tan(2x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of tan(2x) using the first principle, we will consider f(x) = tan(2x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = tan(2x), we write f(x + h) = tan(2(x + h)). Substituting these into equation (1), f'(x) = limₕ→₀ [tan(2(x + h)) - tan(2x)] / h = limₕ→₀ [ [sin(2(x + h)) / cos(2(x + h))] - [sin(2x) / cos(2x)] ] / h = limₕ→₀ [ [sin(2x + 2h) cos(2x) - cos(2x + 2h) sin(2x)] / [cos(2x) · cos(2(x + h))] ] / h We now use the formula sin A cos B - cos A sin B = sin(A - B). f'(x) = limₕ→₀ [ sin(2h) ] / [ h cos(2x) · cos(2(x + h))] = limₕ→₀ [ sin(2h) ] / [ h cos(2x) · cos(2(x + h))] = limₕ→₀ (sin(2h)) / h · limₕ→₀ 1 / [cos(2x) · cos(2(x + h))] Using limit formulas, limₕ→₀ (sin(2h)) / h = 2. f'(x) = 2 [ 1 / (cos(2x) · cos(2x))] = 2/cos²(2x) As the reciprocal of cosine is secant, we have, f'(x) = 2sec²(2x). Hence, proved.

Using Chain Rule

To prove the differentiation of tan(2x) using the chain rule, We use the formula: Tan(2x) = sin(2x)/cos(2x) Consider f(x) = sin(2x) and g(x) = cos(2x) So we get, tan(2x) = 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) = sin(2x) and g(x) = cos(2x) in equation (1), d/dx (tan(2x)) = [(cos(2x) (2cos(2x)) - (sin(2x) (-2sin(2x)))] / (cos(2x))² = (2cos²(2x) + 2sin²(2x)) / cos²(2x) …(2) Here, we use the formula: (cos²(2x)) + (sin²(2x)) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tan(2x)) = 2/ (cos²(2x)) Since sec(2x) = 1/cos(2x), we write: d/dx(tan(2x)) = 2sec²(2x)

Using Product Rule

We will now prove the derivative of tan(2x) using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, Tan(2x) = sin(2x)/cos(2x) tan(2x) = (sin(2x)). (cos(2x))⁻¹ Given that, u = sin(2x) and v = (cos(2x))⁻¹ Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (sin(2x)) = 2cos(2x). (substitute u = sin(2x)) Here we use the chain rule: v = (cos(2x))⁻¹ = (cos(2x))⁻¹ (substitute v = (cos(2x))⁻¹) v' = -1. (cos(2x))⁻². d/dx (cos(2x)) v' = 2sin(2x)/ (cos(2x))² Again, use the product rule formula: d/dx (tan(2x)) = u'. v + u. v' Let’s substitute u = sin(2x), u' = 2cos(2x), v = (cos(2x))⁻¹, and v' = 2sin(2x)/ (cos(2x))² When we simplify each term: We get, d/dx (tan(2x)) = 2 + 2sin²x / (cos(2x))² Sin²x / (cos(2x))² = tan²(2x) (we use the identity sin²x + cos²x =1) Thus: d/dx (tan(2x)) = 2 + 2tan²(2x) Since, 2 + 2tan²(2x) = 2sec²(2x) d/dx (tan(2x)) = 2sec²(2x).