Last updated on September 6, 2025
We use the derivative of 2sin(x), which is 2cos(x), as a measuring tool for how the function 2sin(x) 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 2sin(x) in detail.
We now understand the derivative of 2sinx.
It is commonly represented as d/dx (2sinx) or (2sinx)', and its value is 2cosx.
The function 2sinx has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Sine Function: sin(x) is a basic trigonometric function.
Derivative of Sine: The derivative of sin(x) is cos(x).
Constant Multiplication Rule: If a function is multiplied by a constant, its derivative is the constant multiplied by the derivative of the function.
The derivative of 2sinx can be denoted as d/dx (2sinx) or (2sinx)'. The formula we use to differentiate 2sinx is: d/dx (2sinx) = 2cosx (or) (2sinx)' = 2cosx The formula applies to all x.
We can derive the derivative of 2sinx 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 the Constant Multiple Rule
Using Chain Rule
We will now demonstrate that the differentiation of 2sinx results in 2cosx using the above-mentioned methods:
By First Principle
The derivative of 2sinx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 2sinx using the first principle, we will consider f(x) = 2sinx.
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = 2sinx, we write f(x + h) = 2sin(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [2sin(x + h) - 2sinx] / h = limₕ→₀ 2[sin(x + h) - sinx] / h = 2 limₕ→₀ [sin(x + h) - sinx] / h
Using the formula for the derivative of sinx, = 2cosx
Hence, proved.
Using Constant Multiplication Rule
To prove the differentiation of 2sinx using the constant multiplication rule,
We use the formula: d/dx [c·f(x)] = c·f'(x) Let c = 2 and f(x) = sinx, so we have, d/dx (2sinx) = 2·d/dx (sinx) = 2cosx
Using Chain Rule
To prove the differentiation of 2sinx using the chain rule, Consider y = 2sinx = 2·sinx
Using the chain rule, dy/dx = 2 · d/dx (sinx) = 2cosx
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 2sin(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 2sin(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 (continuing for higher-order derivatives).
When x is 0 or any multiple of π, the derivative 2cos(x) is not zero since cos(0) = 1, so the derivative is 2. The function 2sinx is continuous and differentiable for all x.
Students frequently make mistakes when differentiating 2sinx. 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 (2sinx·cosx)
Here, we have f(x) = 2sinx·cosx.
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 2sinx and v = cosx.
Let’s differentiate each term, u′ = d/dx (2sinx) = 2cosx v′ = d/dx (cosx) = -sinx
Substituting into the given equation, f'(x) = (2cosx)·(cosx) + (2sinx)·(-sinx)
Let’s simplify terms to get the final answer, f'(x) = 2cos²x - 2sin²x
Thus, the derivative of the specified function is 2cos²x - 2sin²x.
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 pendulum swings with its position given by the function y = 2sin(x), where y represents the displacement from the center. If x = π/6 radians, find the velocity of the pendulum.
We have y = 2sin(x) (position of the pendulum)...(1)
Now, we will differentiate the equation (1)
Take the derivative 2sin(x): dy/dx = 2cos(x)
Given x = π/6 (substitute this into the derivative) dy/dx = 2cos(π/6)
We know that cos(π/6) = √3/2 dy/dx = 2(√3/2) = √3
Hence, we get the velocity of the pendulum at x = π/6 as √3.
We find the velocity of the pendulum at x = π/6 as √3, which means that at that point, the pendulum is moving at a rate of √3 units per unit time.
Derive the second derivative of the function y = 2sin(x).
The first step is to find the first derivative, dy/dx = 2cos(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2cos(x)]
Here we use the derivative of cos(x), which is -sin(x), d²y/dx² = 2(-sin(x)) = -2sin(x)
Therefore, the second derivative of the function y = 2sin(x) is -2sin(x).
We use the step-by-step process, where we start with the first derivative. We then apply the derivative of cos(x) to find the second derivative and simplify the terms to find the final answer.
Prove: d/dx (2sin²(x)) = 4sin(x)cos(x).
Let’s start using the chain rule: Consider y = 2sin²(x) = 2(sin(x))²
To differentiate, we use the chain rule: dy/dx = 2·2sin(x)·d/dx [sin(x)]
Since the derivative of sin(x) is cos(x), dy/dx = 4sin(x)cos(x)
Substituting y = 2sin²(x), d/dx (2sin²(x)) = 4sin(x)cos(x) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(x) with its derivative. As a final step, we substitute y = 2sin²(x) to derive the equation.
Solve: d/dx (2sinx/x)
To differentiate the function, we use the quotient rule: d/dx (2sinx/x) = (d/dx (2sinx)·x - 2sinx·d/dx(x))/x²
We will substitute d/dx (2sinx) = 2cosx and d/dx (x) = 1 = (2cosx·x - 2sinx·1)/x² = (2xcosx - 2sinx)/x² = (2xcosx - 2sinx)/x²
Therefore, d/dx (2sinx/x) = (2xcosx - 2sinx)/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.