Derivative of sin(4x)
2026-02-28 06:02 Diff
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Last updated on September 10, 2025

We use the derivative of sin(4x), which is 4cos(4x), 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(4x) in detail.

What is the Derivative of sin(4x)?

We now understand the derivative of sin(4x). It is commonly represented as d/dx (sin(4x)) or (sin(4x))', and its value is 4cos(4x). The function sin(4x) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Sine Function: (sin(x) is a periodic function).

Chain Rule: Rule for differentiating composite functions like sin(4x).

Cosine Function: cos(x) is the derivative of sin(x).

Derivative of sin(4x) Formula

The derivative of sin(4x) can be denoted as d/dx (sin(4x)) or (sin(4x))'.

The formula we use to differentiate sin(4x) is: d/dx (sin(4x)) = 4cos(4x)

The formula applies to all x.

Proofs of the Derivative of sin(4x)

We can derive the derivative of sin(4x) 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

We will now demonstrate that the differentiation of sin(4x) results in 4cos(4x) using the above-mentioned methods:

By First Principle

The derivative of sin(4x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of sin(4x) using the first principle, we will consider f(x) = sin(4x). 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(4x), we write f(x + h) = sin(4(x + h)).

Substituting these into equation (1), f'(x) = limₕ→₀ [sin(4(x + h)) - sin(4x)] / h = limₕ→₀ [ 2cos(4x + 2h)sin(2h) ] / h Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = 4cos(4x) Hence, proved.

Using Chain Rule

To prove the differentiation of sin(4x) using the chain rule, We use the formula: Let u = 4x Then, d/dx (u) = 4 So, d/dx (sin(u)) = cos(u) · d/dx (u)

Substituting back, d/dx (sin(4x)) = cos(4x) · 4 = 4cos(4x)

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Higher-Order Derivatives of sin(4x)

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(4x).

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(4x), 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 an integer multiple of π/4, the derivative is defined and will be a specific value based on the cosine function. When x = 0, the derivative of sin(4x) = 4cos(0), which is 4.

Common Mistakes and How to Avoid Them in Derivatives of sin(4x)

Students frequently make mistakes when differentiating sin(4x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Problem 1

Calculate the derivative of (sin(4x)·cos(4x))

Okay, lets begin

Here, we have f(x) = sin(4x)·cos(4x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(4x) and v = cos(4x). Let’s differentiate each term, u′= d/dx (sin(4x)) = 4cos(4x) v′= d/dx (cos(4x)) = -4sin(4x) Substituting into the given equation, f'(x) = (4cos(4x))·(cos(4x)) + (sin(4x))·(-4sin(4x)) Let’s simplify terms to get the final answer, f'(x) = 4cos²(4x) - 4sin²(4x) Thus, the derivative of the specified function is 4(cos²(4x) - sin²(4x)).

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.

Well explained 👍

Problem 2

A company is modeling the growth of a plant using the function y = sin(4x), where y represents the height of the plant at time x. If x = π/8 weeks, calculate the rate of growth of the plant.

Okay, lets begin

We have y = sin(4x) (growth of the plant)...(1) Now, we will differentiate the equation (1) Take the derivative sin(4x): dy/dx = 4cos(4x) Given x = π/8 (substitute this into the derivative) 4cos(4(π/8)) = 4cos(π/2) = 4 × 0 = 0 Hence, the rate of growth of the plant at x = π/8 weeks is 0.

Explanation

We find the rate of growth of the plant at x = π/8 weeks as 0, indicating that at this particular time, the height of the plant is not changing.

Well explained 👍

Problem 3

Derive the second derivative of the function y = sin(4x).

Okay, lets begin

The first step is to find the first derivative, dy/dx = 4cos(4x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4cos(4x)] = -16sin(4x) Therefore, the second derivative of the function y = sin(4x) is -16sin(4x).

Explanation

We use the step-by-step process, where we start with the first derivative. We then differentiate again, applying the chain rule, to find the second derivative of the function.

Well explained 👍

Problem 4

Prove: d/dx [(sin(4x))²] = 8sin(4x)cos(4x).

Okay, lets begin

Let’s start using the chain rule: Consider y = (sin(4x))² To differentiate, we use the chain rule: dy/dx = 2sin(4x)·d/dx [sin(4x)] Since the derivative of sin(4x) is 4cos(4x), dy/dx = 2sin(4x)·4cos(4x) = 8sin(4x)cos(4x) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(4x) with its derivative. As a final step, we simplify to derive the equation.

Well explained 👍

Problem 5

Solve: d/dx [sin(4x)/x]

Okay, lets begin

To differentiate the function, we use the quotient rule: d/dx [sin(4x)/x] = (d/dx [sin(4x)]·x - sin(4x)·d/dx(x))/x² We will substitute d/dx [sin(4x)] = 4cos(4x) and d/dx(x) = 1 = (4cos(4x)·x - sin(4x)·1)/x² = (4xcos(4x) - sin(4x))/x² Therefore, d/dx [sin(4x)/x] = (4xcos(4x) - sin(4x))/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.

Well explained 👍

FAQs on the Derivative of sin(4x)

1.Find the derivative of sin(4x).

Using the chain rule for sin(4x) gives: d/dx (sin(4x)) = 4cos(4x)

2.Can we use the derivative of sin(4x) in real life?

Yes, we can use the derivative of sin(4x) in real life in calculating the rate of change of any motion, especially in fields such as biology, physics, and engineering.

3.Is it possible to take the derivative of sin(4x) at the point where x = π/4?

Yes, x = π/4 is a point where sin(4x) is defined, so it is possible to take the derivative at this point.

4.What rule is used to differentiate sin(4x)/x?

We use the quotient rule to differentiate sin(4x)/x: d/dx (sin(4x)/x) = (4xcos(4x) - sin(4x))/x²

5.Are the derivatives of sin(4x) and sin⁻¹x the same?

No, they are different. The derivative of sin(4x) is 4cos(4x), while the derivative of sin⁻¹x is 1/√(1-x²).

Important Glossaries for the Derivative of sin(4x)

  • Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
  • Sine Function: A trigonometric function often represented as sin(x), which describes oscillations and periodic phenomena.
  • Cosine Function: A trigonometric function that is the derivative of sine and is represented as cos(x).
  • Chain Rule: A rule used to differentiate composite functions like sin(kx).
  • Quotient Rule: A rule used to differentiate functions that are the quotient of two other functions.

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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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