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Last updated on September 10, 2025
We use the derivative of 3y, which is 3, as a measuring tool for how the function 3y changes in response to a slight change in y. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 3y in detail.
We now understand the derivative of 3y. It is commonly represented as d/dy (3y) or (3y)', and its value is 3. The function 3y has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Constant Multiplication Rule: If a function is multiplied by a constant, its derivative is the constant multiplied by the derivative of the function.
Function of y: A function that depends on the variable y.
The derivative of 3y can be denoted as d/dy (3y) or (3y)'. The formula we use to differentiate 3y is: d/dy (3y) = 3 (or) (3y)' = 3 The formula applies to all y.
We can derive the derivative of 3y using proofs. To show this, we will use basic differentiation rules. There are several methods we use to prove this, such as:
To prove the differentiation of 3y using the constant multiplication rule, consider that if a constant 'c' multiplies a function f(y), its derivative is c times the derivative of f(y). Thus, if f(y) = y, then d/dy (f(y)) = 1.
Therefore, d/dy (3y) = 3 * d/dy (y) = 3 * 1 = 3.
To find the derivative of 3y using the limit definition, we consider f(y) = 3y. Its derivative can be expressed as the following limit: f'(y) = limₕ→₀ [f(y + h) - f(y)] / h
Given that f(y) = 3y, we write f(y + h) = 3(y + h).
Substituting these into the limit expression, f'(y) = limₕ→₀ [3(y + h) - 3y] / h = limₕ→₀ [3y + 3h - 3y] / h = limₕ→₀ 3h / h = limₕ→₀ 3 = 3.
Hence, proved.
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 3y.
For the first derivative of a function, we write f′(y), 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′′(y). Similarly, the third derivative, f′′′(y), is the result of the second derivative, and this pattern continues.
For the nth Derivative of 3y, we generally use fⁿ(y) for the nth derivative of a function f(y), which tells us the change in the rate of change. (continuing for higher-order derivatives).
Since the function 3y is linear with respect to y, all higher-order derivatives beyond the first derivative are 0. When y = 0, the derivative of 3y = 3, which remains constant at any point on the line.
Students frequently make mistakes when differentiating 3y. 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 (3y + y²).
Here, we have f(y) = 3y + y².
Using the sum rule, f'(y) = d/dy (3y) + d/dy (y²) f'(y) = 3 + 2y
Thus, the derivative of the specified function is 3 + 2y.
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 sum rule to get the final result.
The velocity of a car moving in a straight line is represented by the function v = 3y, where y represents the time in seconds. Calculate the acceleration of the car.
We have v = 3y (velocity of the car)...(1)
Now, we will differentiate the equation (1) Take the derivative of 3y: dv/dy = 3
Since acceleration is the derivative of velocity, the acceleration of the car is 3 m/s².
We find the acceleration of the car by differentiating the velocity function with respect to time. Since the derivative is constant, the car accelerates uniformly.
Derive the second derivative of the function y = 3y.
The first step is to find the first derivative, dy/dy = 3...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dy² = d/dy [3] d²y/dy² = 0
Therefore, the second derivative of the function y = 3y is 0.
We use the step-by-step process, where we start with the first derivative. Since the first derivative is a constant, the second derivative is 0. This pattern continues for higher-order derivatives.
Prove: d/dy (3y²) = 6y.
Let’s start using the power rule: Consider y = 3y²
To differentiate, we use the power rule: dy/dy = 3 * 2y dy/dy = 6y
Substituting y = 3y², d/dy (3y²) = 6y
Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation. We then replace y² with its derivative. As a final step, we substitute y = 3y² to derive the equation.
Solve: d/dy (3y/y).
To differentiate the function, we simplify first: d/dy (3y/y) = d/dy (3)
Since the derivative of a constant is 0, Therefore, d/dy (3y/y) = 0.
In this process, we simplify the given function before differentiating. Since it simplifies to a constant, the derivative is 0.
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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