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Last updated on September 20, 2025
The derivative of te^t highlights how the function te^t changes in response to a small change in t. Derivatives are essential tools in understanding dynamic systems, especially in fields such as economics and engineering. We will now discuss the derivative of te^t in detail.
We now understand the derivative of te^t. It is commonly represented as d/dt (te^t) or (te^t)', and its value is e^t + te^t.
The function te^t has a clearly defined derivative, indicating it is differentiable within its domain. Key concepts include:
Exponential Function: (e^t is the exponential function with base e).
Product Rule: Used for differentiating te^t (since it consists of t multiplied by e^t).
The derivative of te^t can be denoted as d/dt (te^t) or (te^t)'. The formula we use to differentiate te^t is: d/dt (te^t) = e^t + te^t This formula applies to all t in the real number domain.
We can derive the derivative of te^t using proofs. To show this, we will use differentiation rules. Several methods can prove this: -
Let's demonstrate that differentiating te^t results in e^t + te^t using these methods:
The derivative of te^t can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of te^t using the first principle, consider f(t) = te^t. Its derivative can be expressed as the following limit: f'(t) = limₕ→₀ [f(t + h) - f(t)] / h
Given f(t) = te^t, we write f(t + h) = (t + h)e^(t + h).
Substituting these into the equation, f'(t) = limₕ→₀ [(t + h)e^(t + h) - te^t] / h = limₕ→₀ [te^(t + h) + he^(t + h) - te^t] / h = limₕ→₀ [te^(t + h) - te^t + he^(t + h)] / h = limₕ→₀ [t(e^(t + h) - e^t) + he^(t + h)] / h
Using the limit property e^(t + h) = e^t * e^h and the fact limₕ→₀ (e^h - 1)/h = 1, f'(t) = te^t * 1 + e^t = e^t + te^t
To prove the differentiation of te^t using the product rule, Consider f(t) = t and g(t) = e^t so that te^t = f(t)g(t).
By the product rule: d/dt [f(t)g(t)] = f'(t)g(t) + f(t)g'(t) f'(t) = 1 (derivative of t) g'(t) = e^t (derivative of e^t)
Substituting into the product rule, d/dt (te^t) = 1(e^t) + t(e^t) = e^t + te^t
When a function is differentiated multiple times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives offer deeper insights into the behavior of functions.
For instance, consider a vehicle where the velocity changes (first derivative), and the rate at which the velocity changes (second derivative) also varies. Higher-order derivatives help us understand functions like te^t more clearly.
For the first derivative of a function, we denote it as f′(t). The second derivative, f′′(t), is obtained from the first derivative. Similarly, the third derivative, f′′′(t), results from the second derivative, continuing this pattern.
For the nth Derivative of te^t, we generally use f n(t) for the nth derivative of a function f(t), which tells us the change in the rate of change.
- When t is 0, the derivative of te^t = e^0 + 0 * e^0, which is 1. - As t approaches -∞, the derivative approaches 0 since e^t diminishes.
Students frequently make mistakes when differentiating te^t. 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 (te^t + e^t).
Here, we have f(t) = te^t + e^t.
Using the sum rule and product rule, f'(t) = d/dt(te^t) + d/dt(e^t) = (e^t + te^t) + e^t = e^t + te^t + e^t = 2e^t + te^t
We find the derivative of the given function by applying the sum rule and the product rule. The first step is to differentiate each part and then combine the results to get the final derivative.
A company models its revenue growth over time using the function R(t) = te^t. Calculate the rate of revenue growth when t = 2.
We have R(t) = te^t (revenue growth function).
Differentiate R(t): dR/dt = e^t + te^t
Substitute t = 2: dR/dt = e^2 + 2e^2 = 3e^2
We find the rate of revenue growth at t = 2 by differentiating the revenue function and substituting t = 2 into the derivative. This provides the rate at which revenue grows at that point in time.
Derive the second derivative of the function f(t) = te^t.
The first step is to find the first derivative, f'(t) = e^t + te^t
Now differentiate f'(t) to get the second derivative: f''(t) = d/dt(e^t + te^t) = e^t + (e^t + te^t) = 2e^t + te^t
We start with the first derivative of f(t) and then differentiate it again to find the second derivative. Using the product and sum rules, we arrive at the final expression.
Prove: d/dt (t^2e^t) = 2te^t + t^2e^t.
Let’s start using the product rule: Consider y = t^2e^t
To differentiate, apply the product rule:
dy/dt = d/dt(t^2) * e^t + t^2 * d/dt(e^t) = 2t * e^t + t^2 * e^t = 2te^t + t^2e^t
We use the product rule to differentiate t^2e^t. Each term is differentiated separately, and then the results are combined to prove the required expression.
Solve: d/dt (te^t/t).
To differentiate the function, we simplify first: d/dt (te^t/t) = d/dt (e^t)
The derivative of e^t is e^t. Therefore, d/dt (te^t/t) = e^t
In this process, we first simplify the given function to a more straightforward form and then differentiate. The simplification helps avoid unnecessary complexity in differentiation.
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