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112 LearnersLast updated on September 1, 2025
Derivative of 2cosx
We use the derivative of 2cos(x), which is -2sin(x), as a tool to understand how the cosine function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of 2cos(x) in detail.
What is the Derivative of 2cosx?
We now understand the derivative of 2cosx. It is commonly represented as d/dx (2cosx) or (2cosx)', and its value is -2sinx.
The function 2cosx has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Cosine Function: (cos(x) is a basic trigonometric function).
Negative Sine: The derivative involves the sine function with a negative sign.
Coefficient Rule: The derivative of a constant multiplied by a function involves multiplying the constant by the derivative of the function.
Derivative of 2cosx Formula
The derivative of 2cosx can be denoted as d/dx (2cosx) or (2cosx)'.
The formula we use to differentiate 2cosx is: d/dx (2cosx) = -2sinx
The formula applies to all x within the domain of the cosine function.
Proofs of the 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
Higher-Order Derivatives of 2cosx
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 2cos(x). 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 2cos(x), 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).
Special Cases:
When x is π/2, the derivative is -2sin(π/2), which is -2.
When x is 0, the derivative of 2cosx = -2sin(0), which is 0.
Common Mistakes and How to Avoid Them in Derivatives of 2cosx
Students frequently make mistakes when differentiating 2cosx.These mistakes can be resolved by understanding the proper solutions.Here are a few common mistakes and ways to solve them:
Mistake 1
Not simplifying the equation
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 unnecessary, but it is important to avoid errors in the process.
Mistake 2
Ignoring the negative sign
Students may overlook the negative sign when differentiating cosine, leading to incorrect results. It's important to remember that the derivative of cosine involves a negative sine factor.
Mistake 3
Incorrect use of the Chain Rule
While differentiating functions such as 2cos(g(x)), students might misapply the chain rule.
For example, incorrect differentiation: d/dx (2cos(g(x))) = -2sin(g(x)).
To avoid this mistake, remember to multiply by the derivative of the inner function as well.
Correct differentiation: d/dx (2cos(g(x))) = -2sin(g(x)) * g'(x).
Mistake 4
Not writing Constants and Coefficients
There is a common mistake where students forget to multiply the constants placed before cosx.
For example, they might incorrectly write d/dx (5cosx) as -sinx.
Students should check the constants in the terms and ensure they are multiplied properly.
For example, the correct equation is d/dx (5cosx) = -5sinx.
Mistake 5
Not Applying the Product Rule
Students often forget to use the product rule when required.This happens when they don't consider terms as products.
Ensure each term is properly differentiated using the correct rules.
For example: Incorrect: d/dx (2cos(x) * x) = -2sin(x). Correct: d/dx (2cos(x) * x) = -2xsin(x) + 2cos(x).
Examples Using the Derivative of 2cosx
Problem 1
Calculate the derivative of (2cosx * sinx)
Here, we have f(x) = 2cosx * sinx.
Using the product rule, f'(x) = u′v + uv′
In the given equation, u = 2cosx and v = sinx.
Let’s differentiate each term, u′= d/dx (2cosx) = -2sinx v′= d/dx (sinx) = cosx
Substituting into the given equation, f'(x) = (-2sinx) * sinx + (2cosx) * cosx
Let’s simplify terms to get the final answer, f'(x) = -2sin²x + 2cos²x
Thus, the derivative of the specified function is -2sin²x + 2cos²x.
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.
Problem 2
A company monitors the oscillation of a pendulum, which is modeled by the function y = 2cos(x), where y represents displacement at time x. If x = π/3 seconds, measure how fast the displacement is changing.
We have y = 2cos(x) (displacement of the pendulum)...(1)
Now, we will differentiate the equation (1)
Take the derivative 2cos(x): dy/dx = -2sin(x)
Given x = π/3 (substitute this into the derivative) dy/dx = -2sin(π/3)
Since sin(π/3) = √3/2, dy/dx = -2 * √3/2 = -√3
Hence, the displacement of the pendulum is changing at a rate of -√3 meters per second at x = π/3.
Explanation
We find the rate of change of displacement at x = π/3 as -√3, which indicates that the displacement is decreasing at this rate.
Problem 3
Derive the second derivative of the function y = 2cos(x).
The first step is to find the first derivative, dy/dx = -2sin(x)...(1) Now we will differentiate equation (1) to get the
second derivative: d²y/dx² = d/dx [-2sin(x)] = -2 * cos(x)
Therefore, the second derivative of the function y = 2cos(x) is -2cos(x).
Explanation
We use the step-by-step process, where we start with the first derivative. Then, we differentiate -2sin(x) to find the second derivative, resulting in -2cos(x).
Problem 4
Prove: d/dx (2cos²(x)) = -4cos(x)sin(x).
Let’s start using the chain rule: Consider y = 2cos²(x) = 2[cos(x)]²
To differentiate, we use the chain rule: dy/dx = 2 * 2cos(x) * d/dx [cos(x)]
Since the derivative of cos(x) is -sin(x), dy/dx = 4cos(x) * (-sin(x)) = -4cos(x)sin(x)
Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation.Then, we replace cos(x) with its derivative. As a final step, we simplify to derive the equation.
Problem 5
Solve: d/dx (2cosx/x)
To differentiate the function, we use the quotient rule:
d/dx (2cosx/x) = (d/dx (2cosx) * x - 2cosx * d/dx(x))/ x²
We will substitute d/dx (2cosx) = -2sinx and d/dx (x) = 1 = (-2sinx * x - 2cosx * 1) / x² = (-2xsinx - 2cosx) / x² = (-2xsinx - 2cosx) / x²
Therefore, d/dx (2cosx/x) = (-2xsinx - 2cosx) / x²
Explanation
In this process, we differentiate the given function using the quotient rule.As a final step, we simplify the equation to obtain the final result.
FAQs on the Derivative of 2cosx
1.Find the derivative of 2cosx.
2.Can we use the derivative of 2cosx in real life?
3.Is it possible to take the derivative of 2cosx at the point where x = π/2?
4.What rule is used to differentiate 2cosx/x?
5.Are the derivatives of 2cosx and cos²x the same?
Important Glossaries for the Derivative of 2cosx
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Cosine Function: A trigonometric function that represents the x-coordinate of a point on the unit circle.
- Sine Function: A trigonometric function representing the y-coordinate of a point on the unit circle.
- Coefficient Rule: A rule stating that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
- Chain Rule: A method for finding the derivative of a composite function.
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.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
