Last updated on July 21st, 2025
We use the derivative of sin(3x), which is 3cos(3x), as a tool to understand how the function changes with respect to x. Derivatives help us calculate rates of change in various real-life situations. We will now discuss the derivative of sin(3x) in detail.
We now understand the derivative of sin(3x). It is commonly represented as d/dx (sin(3x)) or (sin(3x))', and its value is 3cos(3x).
The function sin(3x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Sine Function: The sine function is a fundamental trigonometric function.
Chain Rule: A rule for differentiating composite functions like sin(3x).
Cosine Function: The derivative of the sine function involves the cosine function.
The derivative of sin(3x) can be denoted as d/dx (sin(3x)) or (sin(3x))'.
The formula we use to differentiate sin(3x) is: d/dx (sin(3x)) = 3cos(3x) (sin(3x))' = 3cos(3x) The formula applies to all x.
We can derive the derivative of sin(3x) 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 Using Product Rule We will now demonstrate that the differentiation of sin(3x) results in 3cos(3x) using the above-mentioned methods:
By First Principle The derivative of sin(3x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of sin(3x) using the first principle, we will consider f(x) = sin(3x).
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(3x), we write f(x + h) = sin(3(x + h)).
Substituting these into equation (1), f'(x) = limₕ→₀ [sin(3(x + h)) - sin(3x)] / h = limₕ→₀ [2cos(3x + 3h/2)sin(3h/2)] / h
Using the limit formula for sin(θ)/θ = 1 as θ approaches 0, f'(x) = 3cos(3x) Hence, proved.
Using Chain Rule To prove the differentiation of sin(3x) using the chain rule, Let g(x) = 3x and f(g) = sin(g). Then, f(x) = sin(3x) = sin(g(x)).
Using the chain rule: d/dx [f(g(x))] = f'(g(x)) * g'(x) f'(x) = cos(3x) * 3 = 3cos(3x) Thus, d/dx (sin(3x)) = 3cos(3x).
Using Product Rule We will now prove the derivative of sin(3x) using the product rule.
The step-by-step process is demonstrated below: We express sin(3x) as a product: sin(3x) = sin(3·x)
Given that, u = sin(3) and v = x Using the product rule formula: d/dx [u·v] = u'v + uv' u' = 0 (since sin(3) is constant) v' = 1 d/dx (sin(3x)) = 0·x + sin(3)·1 Simplifying, using chain rule: d/dx (sin(3x)) = 3cos(3x) Thus, the derivative of sin(3x) is 3cos(3x).
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(3x).
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(3x), 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).
When x is such that sin(3x) is at a maximum or minimum, the derivative is zero because the slope is horizontal at those points. When x is 0, the derivative of sin(3x) = 3cos(0) = 3.
Students frequently make mistakes when differentiating sin(3x).
These mistakes can be resolved by understanding the proper solutions.
Here are a few common mistakes and ways to solve them:
Calculate the derivative of (sin(3x)·cos(3x))
Here, we have f(x) = sin(3x)·cos(3x).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(3x) and v = cos(3x).
Let’s differentiate each term, u′ = d/dx (sin(3x)) = 3cos(3x) v′ = d/dx (cos(3x)) = -3sin(3x)
substituting into the given equation, f'(x) = (3cos(3x))(cos(3x)) + (sin(3x))(-3sin(3x))
Let’s simplify terms to get the final answer, f'(x) = 3cos²(3x) - 3sin²(3x)
Thus, the derivative of the specified function is 3(cos²(3x) - sin²(3x)).
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.
A light is swinging in a circular path, and its height from the ground is given by y = sin(3x), where x is time in seconds. Calculate the rate of change of height when x = π/6 seconds.
We have y = sin(3x) (height of the light)...(1) Now, we will differentiate the equation (1)
Take the derivative sin(3x): dy/dx = 3cos(3x) Given x = π/6 (substitute this into the derivative) dy/dx = 3cos(3(π/6)) dy/dx = 3cos(π/2)
Since cos(π/2) = 0, dy/dx = 0 Hence, the rate of change of height when x = π/6 seconds is 0.
We find the rate of change of height by differentiating the height function. At x = π/6, the height change is zero, indicating the light is momentarily at its highest or lowest point.
Derive the second derivative of the function y = sin(3x).
The first step is to find the first derivative, dy/dx = 3cos(3x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3cos(3x)] d²y/dx² = 3[-3sin(3x)] d²y/dx² = -9sin(3x)
Therefore, the second derivative of the function y = sin(3x) is -9sin(3x).
We use the step-by-step process, where we start with the first derivative.
Using the chain rule, we differentiate 3cos(3x).
We then proceed to find the second derivative by differentiating again.
Prove: d/dx (sin²(3x)) = 6sin(3x)cos(3x).
Let’s start using the chain rule: Consider y = sin²(3x) y = [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) dy/dx = 6sin(3x)cos(3x) Hence proved.
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.
Solve: d/dx (sin(3x)/x)
To differentiate the function, we use the quotient rule: d/dx (sin(3x)/x) = (d/dx (sin(3x)).x - sin(3x).d/dx(x))/x²
We will substitute d/dx (sin(3x)) = 3cos(3x) and d/dx (x) = 1 = (3cos(3x).x - sin(3x)) / x² = (3xcos(3x) - sin(3x)) / x²
Therefore, d/dx (sin(3x)/x) = (3xcos(3x) - sin(3x)) / x²
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.
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.
: He loves to play the quiz with kids through algebra to make kids love it.