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106 LearnersLast updated on September 22, 2025
Derivative of Sin(5x)
We use the derivative of sin(5x), which is 5cos(5x), as a measuring tool for how the sine function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of sin(5x) in detail.
What is the Derivative of Sin(5x)?
We now understand the derivative of sin(5x). It is commonly represented as d/dx (sin(5x)) or (sin(5x))', and its value is 5cos(5x).
The function sin(5x) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Sine Function: sin(5x) is a trigonometric function.
Chain Rule: Rule for differentiating sin(5x) (since it involves a composite function).
Cosine Function: cos(x) is another primary trigonometric function.
Derivative of Sin(5x) Formula
The derivative of sin(5x) can be denoted as d/dx (sin(5x)) or (sin(5x))'. The formula we use to differentiate sin(5x) is: d/dx (sin(5x)) = 5cos(5x) (sin(5x))' = 5cos(5x) The formula applies to all x.
Proofs of the Derivative of Sin(5x)
We can derive the derivative of sin(5x) 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:
- By First Principle
- Using Chain Rule
We will now demonstrate that the differentiation of sin(5x) results in 5cos(5x) using the above-mentioned methods:
By First Principle
The derivative of sin(5x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of sin(5x) using the first principle, we will consider f(x) = sin(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) = sin(5x), we write f(x + h) = sin(5(x + h)).
Substituting these into equation (1), f'(x) = limₕ→₀ [sin(5(x + h)) - sin(5x)] / h = limₕ→₀ [2cos((5x + 5h + 5x)/2)sin(5h/2)] / h = limₕ→₀ [2cos(5x + 5h/2)sin(5h/2)] / h
Using limit formulas, limₕ→₀ (sin(5h/2))/(5h/2) = 5/2. f'(x) = 5cos(5x) Hence, proved.
Using Chain Rule
To prove the differentiation of sin(5x) using the chain rule, We use the formula:
Let u = 5x, then sin(u) = sin(5x) d/dx (sin(5x)) = cos(5x) · d/dx (5x) = cos(5x) · 5 = 5cos(5x).
Higher-Order Derivatives of Sin(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 sin(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 sin(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/2, the derivative is 0 because cos(5x) is 0 at x = π/2. When x is 0, the derivative of sin(5x) = 5cos(0), which is 5.
Common Mistakes and How to Avoid Them in Derivatives of Sin(5x)
Students frequently make mistakes when differentiating sin(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 applying the Chain Rule correctly
Students may forget to apply the chain rule when differentiating sin(5x), which can lead to incorrect results. They often write the derivative as cos(5x) instead of 5cos(5x). Ensure that you multiply by the derivative of the inside function (5x), which is 5.
Mistake 2
Confusing the derivative with the integral
Students might confuse the derivative of sin(5x) with its integral. Remember, differentiation and integration are inverse processes. Ensure you’re following the correct process for the task at hand.
Mistake 3
Incorrect use of trigonometric identities
While differentiating, students may misapply trigonometric identities. For example, they might incorrectly try to use identities like sin²(x) + cos²(x) = 1. Focus on applying the chain rule and getting the correct derivative.
Mistake 4
Not multiplying by the derivative of the inner function
There is a common mistake that students at times forget to multiply by the derivative of the inner function (5x). For example, they incorrectly write d/dx (sin(5x)) = cos(5x). Students should check the inside function’s derivative, which is 5. For example, the correct derivative is 5cos(5x).
Mistake 5
Not considering the angle in radians
Students often forget to consider that the angle should be in radians when working with derivatives of trigonometric functions. Ensure that calculations are done in radians for accurate results.
Examples Using the Derivative of Sin(5x)
Problem 1
Calculate the derivative of (sin(5x)·cos(5x))
Here, we have f(x) = sin(5x)·cos(5x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(5x) and v = cos(5x).
Let’s differentiate each term, u′ = d/dx (sin(5x)) = 5cos(5x) v′ = d/dx (cos(5x)) = -5sin(5x)
Substituting into the given equation, f'(x) = (5cos(5x))(cos(5x)) + (sin(5x))(-5sin(5x))
Let’s simplify terms to get the final answer, f'(x) = 5cos²(5x) - 5sin²(5x)
Thus, the derivative of the specified function is 5cos²(5x) - 5sin²(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 factory monitors temperature changes using the function y = sin(5x), where y represents the temperature at time x. If x = π/4 minutes, calculate the rate of change of temperature.
We have y = sin(5x) (temperature function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of sin(5x): dy/dx = 5cos(5x)
Given x = π/4 (substitute this into the derivative)
dy/dx = 5cos(5(π/4)) = 5cos(5π/4) Since cos(5π/4) = -√2/2, dy/dx = 5(-√2/2) = -5√2/2
Hence, the rate of change of temperature at x = π/4 minutes is -5√2/2.
Explanation
We find the rate of change of temperature at x = π/4 as -5√2/2, which indicates the temperature is decreasing at that specific moment.
Problem 3
Derive the second derivative of the function y = sin(5x).
The first step is to find the first derivative, dy/dx = 5cos(5x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5cos(5x)] = 5[-5sin(5x)] = -25sin(5x)
Therefore, the second derivative of the function y = sin(5x) is -25sin(5x).
Explanation
We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate 5cos(5x). We then simplify the terms to find the final answer.
Problem 4
Prove: d/dx (sin²(5x)) = 10sin(5x)cos(5x).
Let’s start using the chain rule: Consider y = sin²(5x) = [sin(5x)]²
To differentiate, we use the chain rule: dy/dx = 2sin(5x)·d/dx [sin(5x)]
Since the derivative of sin(5x) is 5cos(5x), dy/dx = 2sin(5x)·5cos(5x) = 10sin(5x)cos(5x)
Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(5x) with its derivative. As a final step, we simplify to derive the equation.
Problem 5
Solve: d/dx (sin(5x)/x)
To differentiate the function, we use the quotient rule: d/dx (sin(5x)/x) = (d/dx (sin(5x))⋅x - sin(5x)⋅d/dx(x))/x²
We will substitute d/dx (sin(5x)) = 5cos(5x) and d/dx (x) = 1 = (5cos(5x)⋅x - sin(5x)⋅1)/x² = (5xcos(5x) - sin(5x))/x²
Therefore, d/dx (sin(5x)/x) = (5xcos(5x) - sin(5x))/x²
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.
FAQs on the Derivative of Sin(5x)
1.Find the derivative of sin(5x).
2.Can we use the derivative of sin(5x) in real life?
3.Is it possible to take the derivative of sin(5x) at the point where x = π/2?
4.What rule is used to differentiate sin(5x)/x?
5.Are the derivatives of sin(5x) and sin⁻¹(5x) the same?
6.Can we find the derivative of the sin(5x) formula?
Important Glossaries for the Derivative of Sin(5x)
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Sine Function: The sine function is one of the primary six trigonometric functions and is written as sin(x).
- Cosine Function: Another primary trigonometric function, written as cos(x), representing the adjacent side over the hypotenuse in a right triangle.
- Chain Rule: A fundamental rule in calculus used to differentiate composite functions.
- Quotient Rule: A method for finding the derivative of a function that is the ratio of two differentiable 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.
