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Last updated on July 18th, 2025

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Derivative of Sin ax

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We use the derivative of sin(ax), which is a cos(ax), as a tool for measuring 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(ax) in detail.

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What is the Derivative of Sin ax?

We now understand the derivative of sin(ax). It is commonly represented as d/dx (sin(ax)) or (sin(ax))', and its value is a cos(ax). The function sin(ax) has a clearly defined derivative, indicating it is differentiable within its domain.

 

The key concepts are mentioned below: Sine Function: (sin(ax) = a sin(x)).

 

Chain Rule: Rule for differentiating sin(ax) (since it consists of a composite function).

 

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

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Derivative of Sin ax Formula

The derivative of sin(ax) can be denoted as d/dx (sin(ax)) or (sin(ax))'. The formula we use to differentiate sin(ax) is: d/dx (sin(ax)) = a cos(ax) (sin(ax))' = a cos(ax) The formula applies to all x.

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Proofs of the Derivative of Sin ax

We can derive the derivative of sin(ax) 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 Product Rule

 

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

 

By First Principle

 

The derivative of sin(ax) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of sin(ax) using the first principle, we will consider f(x) = sin(ax). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = sin(ax), we write f(x + h) = sin(a(x + h)).

 

Substituting these into the equation, f'(x) = limₕ→₀ [sin(a(x + h)) - sin(ax)] / h = limₕ→₀ [sin(ax + ah) - sin(ax)] / h Using the trigonometric identity for sin(A + B), f'(x) = limₕ→₀ [sin(ax)cos(ah) + cos(ax)sin(ah) - sin(ax)] / h = limₕ→₀ [sin(ax)(cos(ah) - 1) + cos(ax)sin(ah)] / h

For small h, cos(ah) ≈ 1 and sin(ah) ≈ ah, f'(x) = limₕ→₀ [sin(ax)(0) + cos(ax)(ah)] / h = limₕ→₀ [a cos(ax)] f'(x) = a cos(ax).

 

Hence, proved.

 

Using Chain Rule To prove the differentiation of sin(ax) using the chain rule, We use the formula: Let u = ax. Then, sin(ax) = sin(u).

 

By the chain rule: d/dx [sin(u)] = cos(u) * du/dx Thus, d/dx [sin(ax)] = cos(ax) * a = a cos(ax).

 

Using Product Rule

 

We will now prove the derivative of sin(ax) using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, sin(ax) = sin(ax).1 Given that, u = sin(ax) and v = 1

 

Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (sin(ax)) = a cos(ax) (substitute u = sin(ax)) v' = d/dx (1) = 0

 

Using the product rule formula: d/dx (sin(ax)) = u'.v + u.v' = a cos(ax).1 + sin(ax).0 = a cos(ax).

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Higher-Order Derivatives of Sin ax

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(ax).

 

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(ax), we generally use f n(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).

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Special Cases:

When x is a multiple of π, the derivative of sin(ax) = a cos(aπ), which is 0 since cos(aπ) is 0. When x is 0, the derivative of sin(ax) = a cos(0), which is a.

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Common Mistakes and How to Avoid Them in Derivatives of Sin ax

Students frequently make mistakes when differentiating sin(ax). 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 to apply the Chain Rule

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They might not remember to apply the chain rule when differentiating sin(ax). The chain rule is crucial because of the variable a in ax. It is essential to consider the derivative of the inner function ax, which is a.

Mistake 3

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Incorrect use of Trigonometric Identities

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While differentiating functions such as sin(ax), students might misapply trigonometric identities. For example, using incorrect identities can lead to wrong results. Always verify the identities used and ensure they are applied correctly.

Mistake 4

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Not writing Constants and Coefficients

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There is a common mistake that students sometimes forget to multiply the constant a when differentiating sin(ax). For example, they incorrectly write d/dx (sin(ax)) = cos(ax), forgetting to include the coefficient a. Ensure that all constants are accounted for in the differentiation process.

Mistake 5

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Misunderstanding Higher-Order Derivatives

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Students often misunderstand higher-order derivatives, thinking each derivative is the same. However, each differentiation changes the function further. For example, the second derivative of sin(ax) is -a² sin(ax), not simply a² sin(ax).

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Examples Using the Derivative of Sin ax

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Problem 1

Calculate the derivative of (sin(ax) · cos(ax))

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Here, we have f(x) = sin(ax) · cos(ax).

 

Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(ax) and v = cos(ax).

 

Let’s differentiate each term, u′ = d/dx (sin(ax)) = a cos(ax) v′ = d/dx (cos(ax)) = -a sin(ax)

 

substituting into the given equation, f'(x) = (a cos(ax))(cos(ax)) + (sin(ax))(-a sin(ax)) = a cos²(ax) - a sin²(ax)

 

Thus, the derivative of the specified function is a (cos²(ax) - sin²(ax)).

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

AXB International School sponsors the construction of a bridge. The height is represented by the function y = sin(2x), where y represents the elevation of the bridge at a distance x. If x = π/4 meters, measure the elevation rate of the bridge.

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We have y = sin(2x) (elevation of the bridge)...(1)

 

Now, we will differentiate the equation (1)

 

Take the derivative sin(2x): dy/dx = 2 cos(2x)

 

Given x = π/4 (substitute this into the derivative)

 

dy/dx = 2 cos(2(π/4)) = 2 cos(π/2) = 2 * 0 = 0

 

Hence, the elevation rate of the bridge at a distance x = π/4 is 0.

Explanation

We find that the elevation rate of the bridge at x = π/4 is 0, meaning that at this point, the height of the bridge does not change with respect to the horizontal distance.

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Problem 3

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

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The first step is to find the first derivative, dy/dx = a cos(ax)...(1)

 

Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [a cos(ax)] = -a² sin(ax)

 

Therefore, the second derivative of the function y = sin(ax) is -a² sin(ax).

Explanation

We use the step-by-step process, where we start with the first derivative.

 

Then, we differentiate a cos(ax) to find the second derivative, simplifying the terms to find the final answer.

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Problem 4

Prove: d/dx (sin²(ax)) = 2a sin(ax) cos(ax).

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Let’s start using the chain rule: Consider y = sin²(ax) = [sin(ax)]²

 

To differentiate, we use the chain rule: dy/dx = 2 sin(ax) · d/dx [sin(ax)]

 

Since the derivative of sin(ax) is a cos(ax), dy/dx = 2 sin(ax) · a cos(ax) = 2a sin(ax) cos(ax)

 

Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation.

 

Then, we replace sin(ax) with its derivative. As a final step, we substitute y = sin²(ax) to derive the equation.

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Problem 5

Solve: d/dx (sin(ax)/x)

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To differentiate the function, we use the quotient rule: d/dx (sin(ax)/x) = (d/dx (sin(ax)) · x - sin(ax) · d/dx(x))/x²

 

We will substitute d/dx (sin(ax)) = a cos(ax) and d/dx (x) = 1 = (a cos(ax) · x - sin(ax) · 1) / x² = (ax cos(ax) - sin(ax)) / x²

 

Therefore, d/dx (sin(ax)/x) = (ax cos(ax) - sin(ax)) / 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.

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FAQs on the Derivative of Sin ax

1.Find the derivative of sin(ax).

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

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3.Is it possible to take the derivative of sin(ax) at the point where x is a multiple of π?

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

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

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6.Can we find the derivative of the sin(ax) formula?

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Important Glossaries for the Derivative of Sin ax

  • 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 typically represented as sin(x), and sin(ax) involves an additional constant multiplier a.

 

  • Cosine Function: A trigonometric function that is the derivative of the sine function, represented as cos(x).

 

  • Chain Rule: A rule for differentiating composite functions, essential for functions like sin(ax).

 

  • First Derivative: The initial result of differentiating a function, providing the rate of change of the function.
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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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