Derivative of 2cosx
2026-02-28 11:25 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-2cosx

We can derive the derivative of 2cosx using proofs.

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

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

By First Principle

Using the Chain Rule

Using the Product Rule

We will now demonstrate that the differentiation of 2cosx results in -2sinx using the above-mentioned methods:

By First Principle The derivative of 2cosx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 2cosx using the first principle, we will consider f(x) = 2cosx.

Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 2cosx, we write f(x + h) = 2cos(x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [2cos(x + h) - 2cosx] / h = 2limₕ→₀ [cos(x + h) - cosx] / h

Using the trigonometric identity for cos(A + B) and limit properties, f'(x) = 2 * limₕ→₀ [(-2sin(x + h/2)sin(h/2)) / h] = 2 * (-sinx)

Therefore, f'(x) = -2sinx.

Hence, proved.

Using the Chain Rule To prove the differentiation of 2cosx using the chain rule, Consider f(x) = 2 and g(x) = cosx So, f(x) * g(x) = 2cosx

By the chain rule, d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) Substitute f(x) = 2 and g(x) = cosx, d/dx (2cosx) = 0 * cosx + 2 * (-sinx) = -2sinx

Using the Product Rule We will now prove the derivative of 2cosx using the product rule.

The step-by-step process is demonstrated below: Here, we use the formula, 2cosx = 2 * cosx Let u = 2 and v = cosx

Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (2) = 0 v' = d/dx (cosx) = -sinx Using the product rule formula: d/dx (2cosx) = u'.v + u.v' = 0 * cosx + 2 * (-sinx) = -2sinx