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Last updated on July 22nd, 2025

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Derivative of Cos2t

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We use the derivative of cos(2t), which is -2sin(2t), as a measuring tool for how the cosine function changes in response to a slight change in t. Derivatives help us calculate various rates of change in real-life situations. We will now talk about the derivative of cos(2t) in detail.

Derivative of Cos2t for Indonesian Students
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What is the Derivative of Cos2t?

We now understand the derivative of cos(2t). It is commonly represented as d/dt (cos(2t)) or (cos(2t))', and its value is -2sin(2t). The function cos(2t) has a clearly defined derivative, indicating it is differentiable within its domain.

 

The key concepts are mentioned below:

 

Cosine Function: (cos(2t) is a trigonometric function).

 

Chain Rule: Rule for differentiating composite functions like cos(2t).

 

Sine Function: sin(2t) is the derivative of the cosine function, with a negative sign and a coefficient due to the chain rule application.

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Derivative of Cos2t Formula

The derivative of cos(2t) can be denoted as d/dt (cos(2t)) or (cos(2t))'. The formula we use to differentiate cos(2t) is: d/dt (cos(2t)) = -2sin(2t) (cos(2t))' = -2sin(2t)

 

The formula applies to all t where the function is defined.

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Proofs of the Derivative of Cos2t

We can derive the derivative of cos(2t) 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:

 

  1. By First Principle
  2. Using Chain Rule
  3. Using Trigonometric Identities

 

We will now demonstrate that the differentiation of cos(2t) results in -2sin(2t) using the above-mentioned methods:

 

By First Principle

 

The derivative of cos(2t) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

 

To find the derivative of cos(2t) using the first principle, we will consider f(t) = cos(2t). Its derivative can be expressed as the following limit. f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1) Given that f(t) = cos(2t), we write f(t + h) = cos(2(t + h)).

 

Substituting these into equation (1), f'(t) = limₕ→₀ [cos(2(t + h)) - cos(2t)] / h Using the trigonometric identity for cosine difference: cos A - cos B = -2 sin((A + B) / 2) sin((A - B) / 2) f'(t) = limₕ→₀ [-2 sin(2t + h) sin(h)] / h = limₕ→₀ [-2 sin(2t + h) (sin(h) / h)]

 

Using limit formulas, limₕ→₀ (sin h) / h = 1. f'(t) = -2 sin(2t) Hence, proved.

 

Using Chain Rule

 

To prove the differentiation of cos(2t) using the chain rule, We use the formula: Let u = 2t, thus cos(2t) = cos(u).

 

By the chain rule: d/dt (cos(u)) = -sin(u) du/dt d/dt (cos(2t)) = -sin(2t) (2) = -2sin(2t)

 

Using Trigonometric Identities

 

We can also prove the derivative of cos(2t) using trigonometric identities. Let u = 2t, thus: d/dt (cos(2t)) = d/dt (cos(u)) = -sin(u) (2) = -2sin(2t)

 

Thus, the derivative is -2sin(2t).

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Higher-Order Derivatives of Cos2t

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 cos(2t).

 

For the first derivative of a function, we write f′(t), 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′′(t). Similarly, the third derivative, f′′′(t), is the result of the second derivative, and this pattern continues.

 

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

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Special Cases:

When t is π/2, the derivative is zero because sin(π) = 0. When t is 0, the derivative of cos(2t) = -2sin(0), which is 0.

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Common Mistakes and How to Avoid Them in Derivatives of Cos2t

Students frequently make mistakes when differentiating cos(2t). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not simplifying the equation

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Students may forget to simplify the equation, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result, especially when solving using the product or chain rule. Ensure that each step is written in order. Students might think it is awkward, but it is important to avoid errors in the process.

Mistake 2

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Forgetting the Coefficients from Chain Rule

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They might not remember that the chain rule introduces a coefficient when differentiating composite functions like cos(2t). Keep in mind that you should consider the derivative of the inner function when differentiating.

Mistake 3

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Incorrect use of Trigonometric Identities

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While differentiating functions involving trigonometric identities, students misapply these identities. For example: Incorrect differentiation: d/dt (cos(2t)) = -sin(2t). The correct chain rule application gives: d/dt (cos(2t)) = -2sin(2t). To avoid this mistake, ensure that the trigonometric identities are applied correctly and the chain rule is used properly.

Mistake 4

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Not writing Constants and Coefficients

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There is a common mistake that students at times forget to multiply the constants placed before cos(2t). For example, they incorrectly write d/dt (5 cos(2t)) = -sin(2t). Students should check the constants in the terms and ensure they are multiplied properly. For e.g., the correct equation is d/dt (5 cos(2t)) = -10 sin(2t).

Mistake 5

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Not Applying the Chain Rule

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Students often forget to use the chain rule. This happens when the derivative of the inner function is not considered. For example: Incorrect: d/dt (cos(4t)) = -sin(4t). To fix this error, students should divide the functions into inner and outer parts. Then, make sure that each function is differentiated. For example, d/dt (cos(4t)) = -4sin(4t).

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Examples Using the Derivative of Cos2t

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Problem 1

Calculate the derivative of (cos(2t) · sin(2t))

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Here, we have f(t) = cos(2t) · sin(2t). Using the product rule, f'(t) = u′v + uv′ In the given equation, u = cos(2t) and v = sin(2t).

 

Let’s differentiate each term, u′= d/dt (cos(2t)) = -2sin(2t) v′= d/dt (sin(2t)) = 2cos(2t)

 

substituting into the given equation, f'(t) = (-2sin(2t)) · (sin(2t)) + (cos(2t)) · (2cos(2t))

 

Let’s simplify terms to get the final answer, f'(t) = -2sin²(2t) + 2cos²(2t)

 

Thus, the derivative of the specified function is 2cos²(2t) - 2sin²(2t).

Explanation

We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.

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Problem 2

A pendulum swings along an arc, where its displacement from the vertical is given by y = cos(2t). If t = π/3 seconds, what is the rate of change of its displacement?

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We have y = cos(2t) (displacement of the pendulum)...(1)

 

Now, we will differentiate the equation (1).

 

Take the derivative cos(2t): dy/dt = -2sin(2t)

 

Given t = π/3, dy/dt = -2sin(2(π/3)) = -2sin(2π/3) = -2(√3/2)

 

(since sin(2π/3) = √3/2) = -√3

 

Hence, the rate of change of displacement at t = π/3 is -√3 units per second.

Explanation

We find the rate of change of the pendulum's displacement at t = π/3 as -√3, which indicates that the displacement is decreasing at that rate.

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Problem 3

Derive the second derivative of the function y = cos(2t).

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The first step is to find the first derivative, dy/dt = -2sin(2t)...(1)

 

Now we will differentiate equation (1) to get the second derivative:

 

d²y/dt² = d/dt [-2sin(2t)] = -2 [d/dt(sin(2t))] = -2 [2cos(2t)] = -4cos(2t)

 

Therefore, the second derivative of the function y = cos(2t) is -4cos(2t).

Explanation

We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate sin(2t). We then substitute the result and simplify the terms to find the final answer.

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Problem 4

Prove: d/dt (cos²(2t)) = -4cos(2t)sin(2t).

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Let’s start using the chain rule: Consider y = cos²(2t) = [cos(2t)]²

 

To differentiate, we use the chain rule: dy/dt = 2cos(2t) · d/dt [cos(2t)]

 

Since the derivative of cos(2t) is -2sin(2t), dy/dt = 2cos(2t) ·(-2sin(2t)) = -4cos(2t)sin(2t)

 

Substituting y = cos²(2t), d/dt (cos²(2t)) = -4cos(2t)sin(2t)

 

Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace cos(2t) with its derivative. As a final step, we substitute y = cos²(2t) to derive the equation.

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Problem 5

Solve: d/dt (cos(2t)/t)

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To differentiate the function, we use the quotient rule: d/dt (cos(2t)/t) = (d/dt (cos(2t)) · t - cos(2t) · d/dt(t))/t² We will

 

substitute d/dt (cos(2t)) = -2sin(2t) and d/dt(t) = 1 = (-2sin(2t) · t - cos(2t) · 1) / t² = (-2t sin(2t) - cos(2t)) / t²

 

Therefore, d/dt (cos(2t)/t) = (-2t sin(2t) - cos(2t)) / t²

Explanation

In this process, we differentiate the given function using the product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.

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FAQs on the Derivative of Cos2t

1.Find the derivative of cos(2t).

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2.Can we use the derivative of cos(2t) in real life?

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3.Is it possible to take the derivative of cos(2t) at the point where t = π/2?

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4.What rule is used to differentiate cos(2t)/t?

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5.Are the derivatives of cos(2t) and cos⁻¹(t) the same?

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6.Can we find the derivative of the cos(2t) formula?

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Important Glossaries for the Derivative of Cos2t

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

 

  • Cosine Function: The cosine function is one of the primary trigonometric functions and is written as cos(2t).

 

  • Sine Function: A trigonometric function that is a component in the derivative of the cosine function.

 

  • Chain Rule: A rule for differentiating composite functions.

 

  • Trigonometric Identity: An equation involving trigonometric functions that is true for every value of the occurring variables.
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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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