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111 LearnersLast updated on September 22, 2025
Derivative of cos(5x)
We use the derivative of cos(5x), which is -5sin(5x), as a tool to measure 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 cos(5x) in detail.
What is the Derivative of cos(5x)?
We now understand the derivative of cos(5x). It is commonly represented as d/dx (cos(5x)) or (cos(5x))', and its value is -5sin(5x). The function cos(5x) 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 side over the hypotenuse in a right triangle).
Chain Rule: A rule for differentiating composite functions, which is applied to cos(5x).
Sine Function: sin(x) is the opposite side over the hypotenuse in a right triangle.
Derivative of cos(5x) Formula
The derivative of cos(5x) can be denoted as d/dx (cos(5x)) or (cos(5x))'.
The formula we use to differentiate cos(5x) is: d/dx (cos(5x)) = -5sin(5x) (or) (cos(5x))' = -5sin(5x)
The formula applies to all x where the function is defined.
Proofs of the Derivative of cos(5x)
We can derive the derivative of cos(5x) 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
- Using Product Rule
We will now demonstrate that the differentiation of cos(5x) results in -5sin(5x) using the above-mentioned methods:
By First Principle
The derivative of cos(5x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of cos(5x) using the first principle, we will consider f(x) = cos(5x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = cos(5x), we write f(x + h) = cos(5(x + h)).
Substituting these into equation (1), f'(x) = limₕ→₀ [cos(5(x + h)) - cos(5x)] / h = limₕ→₀ [cos(5x + 5h) - cos(5x)] / h = limₕ→₀ [-2sin(5x + 5h/2)sin(5h/2)] / h = limₕ→₀ [-2sin(5x + 5h/2)sin(5h/2)] / (h/2) · (1/2)
Using limit formulas, limₕ→₀ sin(5h/2) / (5h/2) = 1/5, f'(x) = -5sin(5x)
Hence, proved.
Using Chain Rule
To prove the differentiation of cos(5x) using the chain rule, We use the formula: Let u = 5x, then cos(u) By chain rule: d/dx [cos(u)] = -sin(u) · du/dx … (1)
Let’s substitute u = 5x in equation (1), d/dx (cos(5x)) = -sin(5x) · 5 = -5sin(5x)
Hence, proved.
Using Product Rule
We will now prove the derivative of cos(5x) using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, Let u = cos(x) and v = 5, then cos(5x) = cos(x · 5)
Using the product rule formula: d/dx [u · v] = u' · v + u · v' u' = d/dx (cos(x)) = -sin(x). (substitute u = cos(x))
Here we use the chain rule: v = 5 v' = 0
Again, use the product rule formula: d/dx (cos(5x)) = (-sin(x)) · 5 + cos(x) · 0 = -5sin(x)
Thus: d/dx (cos(5x)) = -5sin(5x)
Higher-Order Derivatives of cos(5x)
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(5x).
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 cos(5x), 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/10, the derivative is -5sin(π/2), which is -5. When x is 0, the derivative of cos(5x) = -5sin(0), which is 0.
Common Mistakes and How to Avoid Them in Derivatives of cos(5x)
Students frequently make mistakes when differentiating cos(5x). 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 awkward, but it is important to avoid errors in the process.
Mistake 2
Forgetting the Undefined Points of cos(5x)
They might not remember that cos(5x) is undefined at points where x leads to undefined behavior of the function it's part of. Keep in mind that you should consider the domain of the function that you differentiate. It will help you understand that the function might not be continuous at certain points.
Mistake 3
Incorrect use of Chain Rule
While differentiating functions such as cos(5x), students might misapply the chain rule. For example: Incorrect differentiation: d/dx (cos(5x)) = -sin(5x). d/dx (u) = -sin(u) · du/dx (where u = 5x) Applying the chain rule, d/dx (cos(5x)) = -sin(5x) · 5 To avoid this mistake, write the chain rule without errors. Always check for errors in the calculation and ensure it is properly simplified.
Mistake 4
Not writing Constants and Coefficients
There is a common mistake that students at times forget to multiply the constants placed before cos(5x). For example, they incorrectly write d/dx (3cos(5x)) = -sin(5x). Students should check the constants in the terms and ensure they are multiplied properly. For example, the correct equation is d/dx (3cos(5x)) = -3(5sin(5x)).
Mistake 5
Not Applying the Chain Rule
Students often forget to use the chain rule. This happens when the derivative of the inner function is not considered. For example: Incorrect: d/dx (cos(3x)) = -sin(3x). 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/dx (cos(3x)) = -3sin(3x).
Examples Using the Derivative of cos(5x)
Problem 1
Calculate the derivative of cos(5x) · sin(5x)
Here, we have f(x) = cos(5x) · sin(5x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = cos(5x) and v = sin(5x).
Let’s differentiate each term, u′= d/dx (cos(5x)) = -5sin(5x) v′= d/dx (sin(5x)) = 5cos(5x)
Substituting into the given equation, f'(x) = (-5sin(5x)) · (sin(5x)) + (cos(5x)) · (5cos(5x))
Let’s simplify terms to get the final answer, f'(x) = -5sin²(5x) + 5cos²(5x)
Thus, the derivative of the specified function is 5(cos²(5x) - sin²(5x)).
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 pendulum swings back and forth, and its displacement from the central position at time t is modeled by the function y = cos(5t). If t = π/5 seconds, calculate the rate of change of displacement.
We have y = cos(5t) (displacement of the pendulum)...(1)
Now, we will differentiate the equation (1)
Take the derivative of cos(5t): dy/dt = -5sin(5t) Given t = π/5 (substitute this into the derivative)
dy/dt = -5sin(5(π/5)) = -5sin(π) = 0
Hence, the rate of change of displacement at t = π/5 seconds is 0.
Explanation
The rate of change of displacement at t = π/5 is 0, indicating that the pendulum is momentarily at rest at this point.
Problem 3
Derive the second derivative of the function y = cos(5x).
The first step is to find the first derivative, dy/dx = -5sin(5x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-5sin(5x)]
Here we use the chain rule, d²y/dx² = -5 · 5cos(5x) = -25cos(5x)
Therefore, the second derivative of the function y = cos(5x) is -25cos(5x).
Explanation
We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate -5sin(5x). We then simplify the terms to find the final answer.
Problem 4
Prove: d/dx (cos²(5x)) = -10cos(5x)sin(5x).
Let’s start using the chain rule: Consider y = cos²(5x) = [cos(5x)]²
To differentiate, we use the chain rule: dy/dx = 2cos(5x) · d/dx [cos(5x)]
Since the derivative of cos(5x) is -5sin(5x), dy/dx = 2cos(5x) · (-5sin(5x)) = -10cos(5x)sin(5x)
Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace cos(5x) with its derivative. As a final step, we simplify to derive the equation.
Problem 5
Solve: d/dx (cos(5x)/x)
To differentiate the function, we use the quotient rule: d/dx (cos(5x)/x) = (d/dx (cos(5x)) · x - cos(5x) · d/dx(x))/ x²
We will substitute d/dx (cos(5x)) = -5sin(5x) and d/dx (x) = 1 = (-5sin(5x) · x - cos(5x) · 1) / x² = (-5xsin(5x) - cos(5x)) / x² = -(5xsin(5x) + cos(5x)) / x²
Therefore, d/dx (cos(5x)/x) = -(5xsin(5x) + cos(5x)) / 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 cos(5x)
1.Find the derivative of cos(5x).
2.Can we use the derivative of cos(5x) in real life?
3.Is it possible to take the derivative of cos(5x) at a point where x leads to undefined behavior?
4.What rule is used to differentiate cos(5x)/x?
5.Are the derivatives of cos(5x) and cos⁻¹(x) the same?
6.Can we find the derivative of the cos(5x) formula?
Important Glossaries for the Derivative of cos(5x)
- 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 ratio of the adjacent side to the hypotenuse in a right triangle.
- Sine Function: A trigonometric function that represents the ratio of the opposite side to the hypotenuse in a right triangle.
- Chain Rule: A fundamental rule in calculus used to differentiate composite functions.
- Quotient Rule: A rule used to differentiate the division of two functions.
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.
