Derivative of xe^2x
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Last updated on August 5, 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.

What is the Derivative of xe^2x?

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.

Derivative of xe^2x Formula

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.

Proofs of the Derivative of xe^2x

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.

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Higher-Order Derivatives of xe^2x

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).

Special Cases:

When x is 0, the derivative of xe^2x is (2x + 1)e^2x, which simplifies to e^0 = 1.

Common Mistakes and How to Avoid Them in Derivatives of xe^2x

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:

Problem 1

Calculate the derivative of (xe^2x·ln x)

Okay, lets begin

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.

Explanation

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.

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Problem 2

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.

Okay, lets begin

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.

Explanation

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.

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Problem 3

Derive the second derivative of the function y = xe^2x.

Okay, lets begin

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.

Explanation

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.

Well explained 👍

Problem 4

Prove: d/dx ((xe^2x)²) = 2xe^2x(2x + 1)e^2x.

Okay, lets begin

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.

Explanation

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.

Well explained 👍

Problem 5

Solve: d/dx (xe^2x/x)

Okay, lets begin

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.

Explanation

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.

Well explained 👍

FAQs on the Derivative of xe^2x

1.Find the derivative of xe^2x.

Using the product rule on xe^2x gives x·e^2x, d/dx (xe^2x) = (2x + 1)e^2x (simplified).

2.Can we use the derivative of xe^2x in real life?

Yes, we can use the derivative of xe^2x in real life in calculating the rate of growth or decay, especially in fields such as finance, biology, and physics.

3.Is it possible to take the derivative of xe^2x at any point?

Yes, the function xe^2x is differentiable at all points, so it is possible to take the derivative at any point.

4.What rule is used to differentiate xe^2x·ln x?

We use the product rule to differentiate xe^2x·ln x, d/dx (xe^2x·ln x) = [(2x + 1)e^2x]·ln x + e^2x.

5.Are the derivatives of xe^2x and e^2x the same?

No, they are different. The derivative of xe^2x is (2x + 1)e^2x, while the derivative of e^2x is 2e^2x.

6.Can we find the derivative of the xe^2x formula?

To find, consider y = xe^2x. We use the product rule: y' = [1·e^2x + x·2e^2x] = (2x + 1)e^2x.

Important Glossaries for the Derivative of xe^2x

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.

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Jaskaran Singh Saluja

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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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