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Last updated on September 12, 2025
We use the derivative of t/2, which is 1/2, as a measuring tool for how the function changes in response to a slight change in t. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of t/2 in detail.
We now understand the derivative of t/2. It is commonly represented as d/dt (t/2) or (t/2)', and its value is 1/2. The function t/2 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Linear Function: A function of the form f(t) = at + b, where a and b are constants.
Constant Rule: The derivative of a constant is zero.
Constant Multiple Rule: The derivative of a constant times a function is the constant times the derivative of the function.
The derivative of t/2 can be denoted as d/dt (t/2) or (t/2)'.
The formula we use to differentiate t/2 is: d/dt (t/2) = 1/2 (or) (t/2)' = 1/2
The formula applies to all t.
We can derive the derivative of t/2 using proofs. To show this, we will use basic differentiation rules. There are several methods we use to prove this, such as:
The derivative of t/2 can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of t/2 using the first principle, we will consider f(t) = t/2. Its derivative can be expressed as the following limit.
f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1)
Given that f(t) = t/2, we write f(t + h) = (t + h)/2.
Substituting these into equation (1),
f'(t) = limₕ→₀ [(t + h)/2 - t/2] / h = limₕ→₀ [h/2] / h = limₕ→₀ 1/2 f'(t) = 1/2 Hence, proved.
To prove the differentiation of t/2 using the constant multiple rule, We use the formula: If f(t) = t, then d/dt (f(t)) = 1. So, for f(t) = t/2,
we have: f'(t) = (1/2) * d/dt (t) f'(t) = (1/2) * 1 f'(t) = 1/2
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 t/2.
For the first derivative of a function, we write f′(t), 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′′(t). Similarly, the third derivative, f′′′(t) is the result of the second derivative and this pattern continues.
For the nth Derivative of t/2, we generally use fⁿ(t) for the nth derivative of a function f(t), which tells us the change in the rate of change. (continuing for higher-order derivatives).
When the t is 0, the derivative of t/2 is 1/2.
Students frequently make mistakes when differentiating t/2. 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 (t/2) * 3t
Here, we have f(t) = (t/2) * 3t.
Using the product rule, f'(t) = u′v + uv′ In the given equation, u = t/2 and v = 3t.
Let’s differentiate each term, u′ = d/dt (t/2) = 1/2 v′ = d/dt (3t) = 3
Substituting into the given equation, f'(t) = (1/2) * (3t) + (t/2) * 3
Let’s simplify terms to get the final answer, f'(t) = (3t/2) + (3t/2) f'(t) = 3t
Thus, the derivative of the specified function is 3t.
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 company uses a conveyor belt system where the speed is represented by the function y = t/2, where y represents the speed at time t. If t = 4 seconds, calculate the change in speed.
We have y = t/2 (speed of the conveyor belt)...(1)
Now, we will differentiate the equation (1) Take the derivative of t/2: dy/dt = 1/2
Given t = 4 (substitute this into the derivative) dy/dt = 1/2
Hence, the change in speed at t = 4 seconds is 1/2 units per second.
We find that the change in speed at t = 4 seconds is constant at 1/2, indicating that the speed increases at a constant rate over time.
Derive the second derivative of the function y = t/2.
The first step is to find the first derivative, dy/dt = 1/2...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dt² = d/dt [1/2]
The derivative of a constant is 0.
Therefore, the second derivative of the function y = t/2 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.
Prove: d/dt ((t/2)²) = t/2.
Let’s start using the chain rule: Consider y = (t/2)²
To differentiate, we use the chain rule: dy/dt = 2(t/2) * d/dt(t/2)
Since the derivative of t/2 is 1/2, dy/dt = 2(t/2) * 1/2 dy/dt = t/2
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. We then replace t/2 with its derivative. As a final step, we simplify the expression to derive the equation.
Solve: d/dt ((t/2)/t)
To differentiate the function, we use the quotient rule: d/dt ((t/2)/t) = (d/dt (t/2) * t - (t/2) * d/dt(t)) / t²
We will substitute d/dt (t/2) = 1/2 and d/dt (t) = 1 = (1/2 * t - (t/2) * 1) / t² = (t/2 - t/2) / t² = 0/t²
Therefore, d/dt ((t/2)/t) = 0.
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result, which 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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