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Last updated on July 22nd, 2025

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Derivative of 2xe^-x

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We use the derivative of 2xe^-x to understand how this function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 2xe^-x in detail.

Derivative of 2xe^-x for UK Students
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What is the Derivative of 2xe^-x?

We now understand the derivative of 2xe^-x. It is commonly represented as d/dx (2xe^-x) or (2xe^-x)', and its value is e^-x(2 - 2x). The function 2xe^-x has a clearly defined derivative, indicating it is differentiable within its domain.

 

The key concepts are mentioned below:

 

Exponential Function: (e^-x is an exponential function).

 

Product Rule: Rule for differentiating 2xe^-x (since it consists of 2x multiplied by e^-x).

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Derivative of 2xe^-x Formula

The derivative of 2xe^-x can be denoted as d/dx (2xe^-x) or (2xe^-x)'. The formula we use to differentiate 2xe^-x is: d/dx (2xe^-x) = e^-x(2 - 2x)

 

The formula applies to all x.

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Proofs of the Derivative of 2xe^-x

We can derive the derivative of 2xe^-x using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as: Using Product Rule

 

Using Product Rule

 

We will now prove the derivative of 2xe^-x using the product rule.

 

The step-by-step process is demonstrated below:

 

Let u = 2x and v = e^-x

 

Using the product rule formula:

 

d/dx [u.v] = u'.v + u.v' u' = d/dx (2x) = 2. v' = d/dx (e^-x) = -e^-x.

 

d/dx (2xe^-x) = 2.e^-x + 2x(-e^-x) = 2e^-x - 2xe^-x = e^-x(2 - 2x)

 

Thus, the derivative of 2xe^-x is e^-x(2 - 2x).

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Higher-Order Derivatives of 2xe^-x

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 2xe^-x.

 

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 2xe^-x, we generally use f^n(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).

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Special Cases:

When x = 0, the derivative of 2xe^-x = e^0(2 - 2(0)), which is 2. When x → ∞, the derivative approaches 0 because e^-x approaches 0 faster than any polynomial.

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

Students frequently make mistakes when differentiating 2xe^-x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not applying the Product Rule correctly

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Students may forget to apply the product rule correctly, which can lead to incomplete or incorrect results. Ensure that each term is differentiated separately and then combined as per the product rule. This is crucial to avoid errors in the calculation process.

Mistake 2

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Incorrect Exponential Derivative

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Students might not correctly differentiate the e^-x term. Remember that d/dx(e^-x) = -e^-x. Misapplying this can lead to wrong results. Always keep the rules for differentiating exponential functions in mind.

Mistake 3

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Forgetting to factor out e^-x

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After applying the product rule, students often forget to factor out e^-x, which simplifies the expression. This step is important to present the derivative in its simplest form.

Mistake 4

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Not Evaluating Limits Properly

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When working with limits, such as x → ∞, students might not properly evaluate how e^-x behaves. Remember that as x → ∞, e^-x approaches 0, which affects the behavior of the derivative.

Mistake 5

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Ignoring the Sign of the Derivative

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Students sometimes ignore the negative sign that comes from differentiating e^-x. This can lead to incorrect expressions for the derivative. Always pay attention to signs when differentiating exponential terms.

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Examples Using the Derivative of 2xe^-x

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Problem 1

Calculate the derivative of (2xe^-x · e^x)

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Here, we have f(x) = 2xe^-x · e^x. Using the simplification, 2xe^-x · e^x = 2x.

 

The derivative of 2x is straightforward: f'(x) = d/dx(2x) = 2.

 

Thus, the derivative of the specified function is 2.

Explanation

We find the derivative of the given function by first simplifying the expression, then differentiating the simplified function.

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Problem 2

A company models its revenue with the function R(x) = 2xe^-x, where x represents time in years. Calculate the rate of change of revenue at x = 1 year.

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We have R(x) = 2xe^-x (rate of change of revenue)...(1)

 

Now, we will differentiate the equation (1): dR/dx = e^-x(2 - 2x)

 

Given x = 1 (substitute this into the derivative),

 

dR/dx = e^-1(2 - 2(1)) = e^-1(2 - 2) = 0.

 

Hence, the rate of change of revenue at x = 1 year is 0.

Explanation

We find the rate of change of revenue at x = 1 year as 0, which means that at this point, the revenue is not changing with respect to time.

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Problem 3

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

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The first step is to find the first derivative, dy/dx = e^-x(2 - 2x)...(1)

 

Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx[e^-x(2 - 2x)]

 

Using the product rule and chain rule, d²y/dx² = -e^-x(2 - 2x) + e^-x(-2) = -e^-x(2 - 2x + 2) = -e^-x(4 - 2x).

 

Therefore, the second derivative of the function y = 2xe^-x is -e^-x(4 - 2x).

Explanation

We use the step-by-step process, where we start with the first derivative. Using the product rule and chain rule, we differentiate the expression again to find the second derivative.

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Problem 4

Prove: d/dx (4xe^-x) = e^-x(4 - 4x).

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Let’s start using the product rule: Consider y = 4xe^-x

 

To differentiate, we use the product rule: dy/dx = 4.e^-x + 4x(-e^-x) = 4e^-x - 4xe^-x = e^-x(4 - 4x)

 

Hence proved.

Explanation

In this step-by-step process, we used the product rule to differentiate the equation. Then, we simplify the terms to derive the expression.

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Problem 5

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

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To differentiate the function, we use the quotient rule: d/dx (2xe^-x/x) = (d/dx (2xe^-x) · x - 2xe^-x · d/dx(x))/x²

 

We will substitute d/dx (2xe^-x) = e^-x(2 - 2x) and d/dx (x) = 1, = (e^-x(2 - 2x) · x - 2xe^-x · 1) / x² = (xe^-x(2 - 2x) - 2xe^-x) / x² = e^-x(2x - 2x² - 2x) / x² = e^-x(-2x²) / x² = -2e^-x

 

Therefore, d/dx (2xe^-x/x) = -2e^-x

Explanation

In this process, we differentiate the given function using the product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.

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FAQs on the Derivative of 2xe^-x

1.Find the derivative of 2xe^-x.

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

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3.Is it possible to take the derivative of 2xe^-x at x = ∞?

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4.What rule is used to differentiate 2xe^-x/x?

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

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Professor Greenline from BrightChamps

Important Glossaries for the Derivative of 2xe^-x

  • Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.

 

  • Exponential Function: A function of the form e^x or e^-x, where e is the base of the natural logarithm.

 

  • Product Rule: A rule used to differentiate products of two functions, given by d/dx[u.v] = u'.v + u.v'.

 

  • Chain Rule: A rule used to differentiate compositions of functions.

 

  • Quotient Rule: A rule used to differentiate quotients of two functions, given by d/dx[u/v] = (v.u' - u.v')/v². ```
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Jaskaran Singh Saluja

About the Author

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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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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