Last updated on July 15th, 2025
We use the derivative of xe^2x, which is (2x + 1)e^2x, as a tool to measure how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of xe^2x in detail.
We now understand the derivative of xe^2x. It is commonly represented as d/dx (xe^2x) or (xe^2x)', and its value is (2x + 1)e^2x. The function xe^2x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Exponential Function: e^x is a base function used in the differentiation of xe^2x. Product Rule: Rule for differentiating xe^2x (since it consists of the product of x and e^2x). Constant Function: The number 2 is a constant in the function e^2x.
The derivative of xe^2x can be denoted as d/dx (xe^2x) or (xe^2x)'. The formula we use to differentiate xe^2x is: d/dx (xe^2x) = (2x + 1)e^2x The formula applies to all x.
We can derive the derivative of xe^2x using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as: Using the Product Rule Using the Chain Rule We will now demonstrate that the differentiation of xe^2x results in (2x + 1)e^2x using the above-mentioned methods: Using the Product Rule To prove the differentiation of xe^2x using the product rule, We use the formula: xe^2x = x·e^2x Consider f(x) = x and g(x) = e^2x By product rule: d/dx [f(x)·g(x)] = f'(x)·g(x) + f(x)·g'(x) Let’s substitute f(x) = x and g(x) = e^2x in the product rule, d/dx (xe^2x) = (1)·e^2x + x·(2e^2x) d/dx (xe^2x) = e^2x + 2xe^2x d/dx (xe^2x) = (2x + 1)e^2x Hence, proved. Using the Chain Rule To prove the differentiation of xe^2x using the chain rule, Let y = xe^2x, then y' = d/dx [x·e^2x] Apply the product rule: y' = e^2x + x·d/dx[e^2x] y' = e^2x + x·(2e^2x) y' = (2x + 1)e^2x 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 xe^2x. 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 xe^2x, 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, the derivative of xe^2x is (2x + 1)e^2x, which simplifies to e^0 = 1.
Students frequently make mistakes when differentiating xe^2x. 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 (xe^2x·ln x)
Here, we have f(x) = xe^2x·ln x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = xe^2x and v = ln x. Let’s differentiate each term, u′ = d/dx (xe^2x) = (2x + 1)e^2x v′ = d/dx (ln x) = 1/x Substituting into the given equation, f'(x) = [(2x + 1)e^2x]·ln x + (xe^2x)·(1/x) Let’s simplify terms to get the final answer, f'(x) = (2x + 1)e^2x·ln x + e^2x Thus, the derivative of the specified function is (2x + 1)e^2x·ln x + e^2x.
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 predicts its revenue growth with the function R(x) = xe^2x, where R represents the revenue at time x. If x = 1 year, calculate the rate of change of revenue.
We have R(x) = xe^2x (revenue growth function)...(1) Now, we will differentiate the equation (1) Take the derivative of xe^2x: dR/dx = (2x + 1)e^2x Given x = 1 (substitute this into the derivative) dR/dx = (2(1) + 1)e^2(1) dR/dx = 3e^2 Hence, at x = 1 year, the rate of change of revenue is 3e^2.
We find the rate of change of revenue at x = 1 year as 3e^2, which means that at a given point, the revenue would increase at this rate.
Derive the second derivative of the function y = xe^2x.
The first step is to find the first derivative, dy/dx = (2x + 1)e^2x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(2x + 1)e^2x] Here we use the product rule, d²y/dx² = (2)e^2x + (2x + 1)·2e^2x d²y/dx² = 2e^2x + 2(2x + 1)e^2x d²y/dx² = (4x + 4 + 2)e^2x d²y/dx² = (4x + 6)e^2x Therefore, the second derivative of the function y = xe^2x is (4x + 6)e^2x.
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate (2x + 1)e^2x. We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx ((xe^2x)²) = 2xe^2x(2x + 1)e^2x.
Let’s start using the chain rule: Consider y = (xe^2x)² To differentiate, we use the chain rule: dy/dx = 2(xe^2x)·d/dx [xe^2x] Since the derivative of xe^2x is (2x + 1)e^2x, dy/dx = 2(xe^2x)·(2x + 1)e^2x Substituting y = (xe^2x)², d/dx ((xe^2x)²) = 2xe^2x(2x + 1)e^2x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace xe^2x with its derivative. As a final step, we substitute y = (xe^2x)² to derive the equation.
Solve: d/dx (xe^2x/x)
To differentiate the function, we simplify it first: xe^2x/x = e^2x Now, we differentiate e^2x: d/dx (e^2x) = 2e^2x Therefore, d/dx (xe^2x/x) = 2e^2x.
In this process, we simplify the given function first and then differentiate the simplified function using the basic derivative rule to obtain the final result.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Exponential Function: A mathematical function involving the constant e, often seen in growth and decay problems. Product Rule: A fundamental rule in calculus used to differentiate products of two functions. Chain Rule: A rule used in calculus to differentiate composite functions. Higher-Order Derivative: A derivative that is obtained by differentiating a function multiple times.
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.