Derivative of sin²x
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Last updated on August 5, 2025

We use the derivative of sin²(x), which involves the chain rule, as a tool for understanding how the 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 sin²(x) in detail.

What is the Derivative of sin²x?

We now understand the derivative of sin²x. It is commonly represented as d/dx (sin²x) or (sin²x)', and its value is 2sin(x)cos(x) or sin(2x). The function sin²x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: - Sine Function: (sin(x) is a trigonometric function). - Chain Rule: Used for differentiating composite functions like sin²(x). - Double Angle Formula: sin(2x) = 2sin(x)cos(x).

Derivative of sin²x Formula

The derivative of sin²x can be denoted as d/dx (sin²x) or (sin²x)'. The formula we use to differentiate sin²x is: d/dx (sin²x) = 2sin(x)cos(x) or (sin²x)' = sin(2x) The formula applies to all x where the sine function is defined.

Proofs of the Derivative of sin²x

We can derive the derivative of sin²x 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 Chain Rule - Using Product Rule We will now demonstrate that the differentiation of sin²x results in 2sin(x)cos(x) using the above-mentioned methods: Using Chain Rule To prove the differentiation of sin²x using the chain rule, Consider y = (sin(x))² Let u = sin(x), then y = u² By the chain rule: dy/dx = 2u du/dx Since du/dx = cos(x), dy/dx = 2sin(x)cos(x) Using the double angle formula, sin(2x) = 2sin(x)cos(x), we have: dy/dx = sin(2x) Hence, proved. Using Product Rule We will now prove the derivative of sin²x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, sin²x = sin(x)·sin(x) Let u = sin(x) and v = sin(x) Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = d/dx (sin x) = cos x v' = d/dx (sin x) = cos x d/dx (sin²x) = cos(x)·sin(x) + sin(x)·cos(x) = 2sin(x)cos(x) Thus, d/dx (sin²x) = sin(2x) Hence, proved.

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Higher-Order Derivatives of sin²x

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 sin²(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 sin²(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.

Special Cases:

- When x is 0, the derivative of sin²x = sin(2·0) = 0. - When x is π/4, the derivative of sin²x = sin(π/2) = 1.

Common Mistakes and How to Avoid Them in Derivatives of sin²x

Students frequently make mistakes when differentiating sin²x. 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 (sin²x·cos(x))

Okay, lets begin

Here, we have f(x) = sin²x·cos(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin²x and v = cos(x). Let’s differentiate each term, u′ = d/dx (sin²x) = 2sin(x)cos(x) v′ = d/dx (cos(x)) = -sin(x) Substituting into the given equation, f'(x) = (2sin(x)cos(x))·(cos(x)) + (sin²x)·(-sin(x)) = 2sin(x)cos²(x) - sin³(x) Thus, the derivative of the specified function is 2sin(x)cos²(x) - sin³(x).

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

A company uses a solar panel to measure sunlight intensity, represented by the function y = sin²(x), where y represents the intensity at angle x. If x = π/6 radians, measure the rate of change of sunlight intensity.

Okay, lets begin

We have y = sin²(x) (intensity of sunlight)...(1) Now, we will differentiate the equation (1) Take the derivative of sin²(x): dy/dx = sin(2x) Given x = π/6 (substitute this into the derivative) sin(2x) = sin(2·π/6) = sin(π/3) = √3/2 Hence, the rate of change of sunlight intensity at x = π/6 is √3/2.

Explanation

We find the rate of change of sunlight intensity at x = π/6 as √3/2, which indicates how rapidly the intensity changes at that angle.

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Problem 3

Derive the second derivative of the function y = sin²(x).

Okay, lets begin

The first step is to find the first derivative, dy/dx = sin(2x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [sin(2x)] d²y/dx² = 2cos(2x) Therefore, the second derivative of the function y = sin²(x) is 2cos(2x).

Explanation

We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate sin(2x). We then substitute the identity and simplify the terms to find the final answer.

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Problem 4

Prove: d/dx (sin²(3x)) = 6sin(3x)cos(3x).

Okay, lets begin

Let’s start using the chain rule: Consider y = sin²(3x) = [sin(3x)]² To differentiate, we use the chain rule: dy/dx = 2sin(3x) d/dx [sin(3x)] Since the derivative of sin(3x) is 3cos(3x), dy/dx = 2sin(3x)·3cos(3x) = 6sin(3x)cos(3x) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(3x) with its derivative. As a final step, we substitute y = sin²(3x) to derive the equation.

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Problem 5

Solve: d/dx (sin²x/x)

Okay, lets begin

To differentiate the function, we use the quotient rule: d/dx (sin²x/x) = (d/dx (sin²x)·x - sin²x·d/dx(x))/x² We will substitute d/dx (sin²x) = 2sin(x)cos(x) and d/dx (x) = 1 = (2sin(x)cos(x)·x - sin²x·1) / x² = (2xsin(x)cos(x) - sin²x) / x² Therefore, d/dx (sin²x/x) = (2xsin(x)cos(x) - sin²x) / 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 sin²x

1.Find the derivative of sin²x.

Using the chain rule for sin²x gives (sin(x))², d/dx (sin²x) = 2sin(x)cos(x) = sin(2x) (simplified)

2.Can we use the derivative of sin²x in real life?

Yes, we can use the derivative of sin²x in real life in calculating the rate of change of any motion, especially in fields such as mathematics, physics, and engineering.

3.Is it possible to take the derivative of sin²x at the point where x = π/2?

Yes, at x = π/2, sin²x = 1, and its derivative sin(2x) = 0.

4.What rule is used to differentiate sin²x/x?

We use the quotient rule to differentiate sin²x/x, d/dx (sin²x/x) = (2xsin(x)cos(x) - sin²x) / x².

5.Are the derivatives of sin²x and sin²⁻¹x the same?

No, they are different. The derivative of sin²x is equal to sin(2x) or 2sin(x)cos(x), while the derivative of sin⁻¹x is 1/√(1-x²).

6.Can we find the derivative of the sin²x formula?

To find, consider y = sin²x. We use the chain rule: y′ = 2sin(x)cos(x) = sin(2x).

Important Glossaries for the Derivative of sin²x

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 trigonometric functions, written as sin(x). Chain Rule: A differentiation rule used for composite functions. Double Angle Formula: A trigonometric identity used to express trigonometric functions of double angles, like sin(2x) = 2sin(x)cos(x). Quotient Rule: A rule for differentiating functions that are divided by one another.

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Jaskaran Singh Saluja

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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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