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2026-01-01
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>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.</p>
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<p>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.</p>
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<h2>What is the Derivative of 2xe^-x?</h2>
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<h2>What is the Derivative of 2xe^-x?</h2>
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<p>We now understand the derivative<a>of</a>2xe^-x. It is commonly represented as d/dx (2xe^-x) or (2xe^-x)', and its value is e^-x(2 - 2x). The<a>function</a>2xe^-x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative<a>of</a>2xe^-x. It is commonly represented as d/dx (2xe^-x) or (2xe^-x)', and its value is e^-x(2 - 2x). The<a>function</a>2xe^-x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The key concepts are mentioned below:</p>
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<p>The key concepts are mentioned below:</p>
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<p>Exponential Function: (e^-x is an exponential function).</p>
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<p>Exponential Function: (e^-x is an exponential function).</p>
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<p>Product Rule: Rule for differentiating 2xe^-x (since it consists of 2x multiplied by e^-x).</p>
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<p>Product Rule: Rule for differentiating 2xe^-x (since it consists of 2x multiplied by e^-x).</p>
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<h2>Derivative of 2xe^-x Formula</h2>
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<h2>Derivative of 2xe^-x Formula</h2>
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<p>The derivative of 2xe^-x can be denoted as d/dx (2xe^-x) or (2xe^-x)'. The<a>formula</a>we use to differentiate 2xe^-x is: d/dx (2xe^-x) = e^-x(2 - 2x)</p>
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<p>The derivative of 2xe^-x can be denoted as d/dx (2xe^-x) or (2xe^-x)'. The<a>formula</a>we use to differentiate 2xe^-x is: d/dx (2xe^-x) = e^-x(2 - 2x)</p>
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<p>The formula applies to all x.</p>
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<p>The formula applies to all x.</p>
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<h2>Proofs of the Derivative of 2xe^-x</h2>
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<h2>Proofs of the Derivative of 2xe^-x</h2>
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<p>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</p>
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<p>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</p>
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<h3>Using Product Rule</h3>
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<h3>Using Product Rule</h3>
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<p>We will now prove the derivative of 2xe^-x using the<a>product</a>rule.</p>
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<p>We will now prove the derivative of 2xe^-x using the<a>product</a>rule.</p>
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<p>The step-by-step process is demonstrated below:</p>
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<p>The step-by-step process is demonstrated below:</p>
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<p>Let u = 2x and v = e^-x</p>
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<p>Let u = 2x and v = e^-x</p>
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<p>Using the product rule formula:</p>
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<p>Using the product rule formula:</p>
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<p>d/dx [u.v] = u'.v + u.v' u' = d/dx (2x) = 2. v' = d/dx (e^-x) = -e^-x.</p>
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<p>d/dx [u.v] = u'.v + u.v' u' = d/dx (2x) = 2. v' = d/dx (e^-x) = -e^-x.</p>
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<p>d/dx (2xe^-x) = 2.e^-x + 2x(-e^-x) = 2e^-x - 2xe^-x = e^-x(2 - 2x)</p>
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<p>d/dx (2xe^-x) = 2.e^-x + 2x(-e^-x) = 2e^-x - 2xe^-x = e^-x(2 - 2x)</p>
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<p>Thus, the derivative of 2xe^-x is e^-x(2 - 2x).</p>
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<p>Thus, the derivative of 2xe^-x is e^-x(2 - 2x).</p>
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<h2>Higher-Order Derivatives of 2xe^-x</h2>
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<h2>Higher-Order Derivatives of 2xe^-x</h2>
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<p>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.</p>
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<p>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.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like 2xe^-x.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like 2xe^-x.</p>
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<p>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.</p>
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<p>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.</p>
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<p>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).</p>
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<p>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).</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>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<a>polynomial</a>.</p>
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<p>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<a>polynomial</a>.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 2xe^-x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 2xe^-x</h2>
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<p>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:</p>
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<p>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:</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Calculate the derivative of (2xe^-x · e^x)</p>
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<p>Calculate the derivative of (2xe^-x · e^x)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = 2xe^-x · e^x. Using the simplification, 2xe^-x · e^x = 2x.</p>
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<p>Here, we have f(x) = 2xe^-x · e^x. Using the simplification, 2xe^-x · e^x = 2x.</p>
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<p>The derivative of 2x is straightforward: f'(x) = d/dx(2x) = 2.</p>
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<p>The derivative of 2x is straightforward: f'(x) = d/dx(2x) = 2.</p>
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<p>Thus, the derivative of the specified function is 2.</p>
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<p>Thus, the derivative of the specified function is 2.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative of the given function by first simplifying the expression, then differentiating the simplified function.</p>
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<p>We find the derivative of the given function by first simplifying the expression, then differentiating the simplified function.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have R(x) = 2xe^-x (rate of change of revenue)...(1)</p>
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<p>We have R(x) = 2xe^-x (rate of change of revenue)...(1)</p>
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<p>Now, we will differentiate the equation (1): dR/dx = e^-x(2 - 2x)</p>
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<p>Now, we will differentiate the equation (1): dR/dx = e^-x(2 - 2x)</p>
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<p>Given x = 1 (substitute this into the derivative),</p>
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<p>Given x = 1 (substitute this into the derivative),</p>
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<p>dR/dx = e^-1(2 - 2(1)) = e^-1(2 - 2) = 0.</p>
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<p>dR/dx = e^-1(2 - 2(1)) = e^-1(2 - 2) = 0.</p>
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<p>Hence, the rate of change of revenue at x = 1 year is 0.</p>
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<p>Hence, the rate of change of revenue at x = 1 year is 0.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = 2xe^-x.</p>
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<p>Derive the second derivative of the function y = 2xe^-x.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>The first step is to find the first derivative, dy/dx = e^-x(2 - 2x)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = e^-x(2 - 2x)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx[e^-x(2 - 2x)]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx[e^-x(2 - 2x)]</p>
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<p>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).</p>
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<p>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).</p>
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<p>Therefore, the second derivative of the function y = 2xe^-x is -e^-x(4 - 2x).</p>
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<p>Therefore, the second derivative of the function y = 2xe^-x is -e^-x(4 - 2x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx (4xe^-x) = e^-x(4 - 4x).</p>
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<p>Prove: d/dx (4xe^-x) = e^-x(4 - 4x).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the product rule: Consider y = 4xe^-x</p>
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<p>Let’s start using the product rule: Consider y = 4xe^-x</p>
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<p>To differentiate, we use the product rule: dy/dx = 4.e^-x + 4x(-e^-x) = 4e^-x - 4xe^-x = e^-x(4 - 4x)</p>
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<p>To differentiate, we use the product rule: dy/dx = 4.e^-x + 4x(-e^-x) = 4e^-x - 4xe^-x = e^-x(4 - 4x)</p>
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<p>Hence proved.</p>
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<p>Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this step-by-step process, we used the product rule to differentiate the equation. Then, we simplify the terms to derive the expression.</p>
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<p>In this step-by-step process, we used the product rule to differentiate the equation. Then, we simplify the terms to derive the expression.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dx (2xe^-x/x)</p>
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<p>Solve: d/dx (2xe^-x/x)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>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</p>
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<p>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</p>
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<p>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</p>
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<p>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</p>
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<p>Therefore, d/dx (2xe^-x/x) = -2e^-x</p>
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<p>Therefore, d/dx (2xe^-x/x) = -2e^-x</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on the Derivative of 2xe^-x</h2>
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<h2>FAQs on the Derivative of 2xe^-x</h2>
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<h3>1.Find the derivative of 2xe^-x.</h3>
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<h3>1.Find the derivative of 2xe^-x.</h3>
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<p>Using the product rule, the derivative of 2xe^-x is e^-x(2 - 2x).</p>
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<p>Using the product rule, the derivative of 2xe^-x is e^-x(2 - 2x).</p>
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<h3>2.Can we use the derivative of 2xe^-x in real life?</h3>
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<h3>2.Can we use the derivative of 2xe^-x in real life?</h3>
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<p>Yes, derivatives like that of 2xe^-x can be used in real life to model phenomena such as population growth or decay, optimizing processes, and more.</p>
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<p>Yes, derivatives like that of 2xe^-x can be used in real life to model phenomena such as population growth or decay, optimizing processes, and more.</p>
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<h3>3.Is it possible to take the derivative of 2xe^-x at x = ∞?</h3>
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<h3>3.Is it possible to take the derivative of 2xe^-x at x = ∞?</h3>
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<p>As x approaches infinity, the derivative approaches 0 because e^-x approaches 0 faster than any polynomial.</p>
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<p>As x approaches infinity, the derivative approaches 0 because e^-x approaches 0 faster than any polynomial.</p>
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<h3>4.What rule is used to differentiate 2xe^-x/x?</h3>
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<h3>4.What rule is used to differentiate 2xe^-x/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate 2xe^-x/x, d/dx (2xe^-x/x) involves differentiating 2xe^-x and x separately and applying the quotient rule.</p>
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<p>We use the<a>quotient</a>rule to differentiate 2xe^-x/x, d/dx (2xe^-x/x) involves differentiating 2xe^-x and x separately and applying the quotient rule.</p>
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<h3>5.Are the derivatives of 2xe^-x and xe^-x the same?</h3>
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<h3>5.Are the derivatives of 2xe^-x and xe^-x the same?</h3>
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<p>No, they are different. The derivative of 2xe^-x is e^-x(2 - 2x), whereas the derivative of xe^-x is e^-x(1 - x).</p>
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<p>No, they are different. The derivative of 2xe^-x is e^-x(2 - 2x), whereas the derivative of xe^-x is e^-x(1 - x).</p>
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<h2>Important Glossaries for the Derivative of 2xe^-x</h2>
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<h2>Important Glossaries for the Derivative of 2xe^-x</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form e^x or e^-x, where e is the base of the natural logarithm.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form e^x or e^-x, where e is the base of the natural logarithm.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two functions, given by d/dx[u.v] = u'.v + u.v'.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two functions, given by d/dx[u.v] = u'.v + u.v'.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate quotients of two functions, given by d/dx[u/v] = (v.u' - u.v')/v². ```</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate quotients of two functions, given by d/dx[u/v] = (v.u' - u.v')/v². ```</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>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.</p>
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<p>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.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>