Derivative of 2cos2x

We now understand the derivative of 2cos2x. It is commonly represented as d/dx (2cos2x) or (2cos2x)', and its value is -4sin2x. The function 2cos2x has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Cosine Function: (cos(x) is the adjacent/hypotenuse in a right triangle).

Chain Rule: Rule for differentiating composite functions like 2cos2x.

Sine Function: sin(x) = opposite/hypotenuse in a right triangle.

The derivative of 2cos2x can be denoted as d/dx (2cos2x) or (2cos2x)'.

The formula we use to differentiate 2cos2x is: d/dx (2cos2x) = -4sin2x

The formula applies to all x where cos(x) is defined.

We can derive the derivative of 2cos2x 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 Chain Rule

We will now demonstrate that the differentiation of 2cos2x results in -4sin2x using the above-mentioned methods:

By First Principle

The derivative of 2cos2x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 2cos2x using the first principle, we will consider f(x) = 2cos2x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 2cos2x, we write f(x + h) = 2cos2(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [2cos2(x + h) - 2cos2x] / h = 2 limₕ→₀ [cos2(x + h) - cos2x] / h Using the identity cos(A + B) = cosAcosB - sinAsinB, f'(x) = 2 limₕ→₀ [cos2xcos2h - sin2xsin2h - cos2x] / h = 2 limₕ→₀ [-sin2xsin2h] / h = 2(-2sin2x) limₕ→₀ (sin2h)/2h Using limit formulas, limₕ→₀ (sin2h)/2h = 1. f'(x) = -4sin2x. Hence, proved.

Using Chain Rule

To prove the differentiation of 2cos2x using the chain rule, We use the formula: Let y = 2cos2x, where u = 2x and v = cosu dy/dx = d/dv (cosu) * d/du (2x) dy/dx = -sin(2x) * 2 dy/dx = -4sin2x Thus, the derivative of 2cos2x is -4sin2x.

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like 2cos2x.

For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x) is the result of the second derivative, and this pattern continues.

For the nth Derivative of 2cos2x, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).

When x is π/2, the derivative is -4sin(π), which is 0.

When x is 0, the derivative of 2cos2x = -4sin(0), which is 0.

• Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.

• Chain Rule: A rule used to differentiate composite functions by differentiating the outer and inner functions.

• Cosine Function: A trigonometric function representing the adjacent/hypotenuse in a right triangle.

• Sine Function: A trigonometric function representing the opposite/hypotenuse in a right triangle.

• Quotient Rule: A rule used to differentiate functions presented as a ratio of two differentiable functions.