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2026-01-01
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<p>Last updated on<strong>October 9, 2025</strong></p>
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<p>Last updated on<strong>October 9, 2025</strong></p>
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<p>We use the derivative of 2xe^x as a tool to measure how the function changes in response to a slight change in x. Derivatives are crucial in calculating various real-life applications, such as rates of change in physical systems. We will now discuss the derivative of 2xe^x in detail.</p>
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<p>We use the derivative of 2xe^x as a tool to measure how the function changes in response to a slight change in x. Derivatives are crucial in calculating various real-life applications, such as rates of change in physical systems. We will now discuss 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 2xe^x + 2e^x. The<a>function</a>2xe^x has a well-defined derivative, indicating it is differentiable across 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 2xe^x + 2e^x. The<a>function</a>2xe^x has a well-defined derivative, indicating it is differentiable across 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 the<a>base</a>of natural<a>logarithms</a>).</p>
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<p>Exponential Function: (e^x is the<a>base</a>of natural<a>logarithms</a>).</p>
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<p>Product Rule: Rule for differentiating 2xe^x (since it consists of a<a>product</a>of 2x and e^x).</p>
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<p>Product Rule: Rule for differentiating 2xe^x (since it consists of a<a>product</a>of 2x and 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) = 2xe^x + 2e^x The formula applies to all x in the<a>real number system</a>.</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) = 2xe^x + 2e^x The formula applies to all x in the<a>real number system</a>.</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 2xex using proofs. To show this, we will use differentiation rules.</p>
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<p>We can derive the derivative of 2xex using proofs. To show this, we will use differentiation rules.</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<h2>Using Product Rule</h2>
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<h2>Using Product Rule</h2>
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<p>Verification by First Principle We will now demonstrate that the differentiation of 2xex results in 2xex + 2ex using the above-mentioned methods:</p>
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<p>Verification by First Principle We will now demonstrate that the differentiation of 2xex results in 2xex + 2ex using the above-mentioned methods:</p>
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<h2>Using Product Rule</h2>
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<h2>Using Product Rule</h2>
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<p>To prove the differentiation of 2xex using the product rule, Consider f(x) = 2x and g(x) = ex So we get, 2xex = f(x)·g(x) By product rule: d/dx [f(x)·g(x)] = f'(x)·g(x) + f(x)·g'(x) Let’s substitute f(x) = 2x and g(x) = e^x, d/dx (2xex) = (2)(ex) + (2x)(e^x) = 2ex + 2xex Thus, d/dx (2xe^x) = 2xe^x + 2e^x</p>
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<p>To prove the differentiation of 2xex using the product rule, Consider f(x) = 2x and g(x) = ex So we get, 2xex = f(x)·g(x) By product rule: d/dx [f(x)·g(x)] = f'(x)·g(x) + f(x)·g'(x) Let’s substitute f(x) = 2x and g(x) = e^x, d/dx (2xex) = (2)(ex) + (2x)(e^x) = 2ex + 2xex Thus, d/dx (2xe^x) = 2xe^x + 2e^x</p>
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<h2>Verification by First Principle</h2>
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<h2>Verification by First Principle</h2>
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<p>The derivative of 2xex can be verified using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 2xex using the first principle, we will consider f(x) = 2xex. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [(f(x + h) - f(x)) / h] Given that f(x) = 2xex, we write f(x + h) = 2(x+h)e(x+h). Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [(2(x + h)e(x + h) - 2xex) / h] = limₕ→₀ [2xe(x + h) + 2he(x + h) - 2xex] / h = limₕ→₀ [2xex(eh - 1) + 2he(x + h)] / h By dividing each<a>term</a>by h, we have: = limₕ→₀ [2xex (eh - 1)/h + 2e(x + h)] Using the limit formula limₕ→₀ (eh - 1)/h = 1, = 2xex (1) + 2ex = 2xex + 2ex Hence, verified.</p>
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<p>The derivative of 2xex can be verified using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 2xex using the first principle, we will consider f(x) = 2xex. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [(f(x + h) - f(x)) / h] Given that f(x) = 2xex, we write f(x + h) = 2(x+h)e(x+h). Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [(2(x + h)e(x + h) - 2xex) / h] = limₕ→₀ [2xe(x + h) + 2he(x + h) - 2xex] / h = limₕ→₀ [2xex(eh - 1) + 2he(x + h)] / h By dividing each<a>term</a>by h, we have: = limₕ→₀ [2xex (eh - 1)/h + 2e(x + h)] Using the limit formula limₕ→₀ (eh - 1)/h = 1, = 2xex (1) + 2ex = 2xex + 2ex Hence, verified.</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 bit complex. To understand them better, consider a car whose 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 2xex.</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 bit complex. To understand them better, consider a car whose 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 2xex.</p>
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<p>For the first derivative of a function, we write f′(x), indicating 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), indicating 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 2xex, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.</p>
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<p>For the nth Derivative of 2xex, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.</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 2xex = 2xex + 2ex = 0 + 2 = 2.</p>
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<p>When x = 0, the derivative of 2xex = 2xex + 2ex = 0 + 2 = 2.</p>
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<p>When x approaches negative infinity, the derivative tends toward 0, as ex approaches 0.</p>
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<p>When x approaches negative infinity, the derivative tends toward 0, as ex approaches 0.</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 2xex. 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 2xex. 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) = 2xex·ex. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 2xe^x and v = e^x. Let’s differentiate each term, u′ = d/dx (2xex) = 2xe^x + 2e^x v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (2xe^x + 2e^x)(e^x) + (2xe^x)(e^x) Let’s simplify terms to get the final answer, f'(x) = 2xe^(2x) + 2e^(2x) + 2xe^(2x) = 4xe^(2x) + 2e^(2x) Thus, the derivative of the specified function is 4xe^(2x) + 2e^(2x).</p>
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<p>Here, we have f(x) = 2xex·ex. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 2xe^x and v = e^x. Let’s differentiate each term, u′ = d/dx (2xex) = 2xe^x + 2e^x v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (2xe^x + 2e^x)(e^x) + (2xe^x)(e^x) Let’s simplify terms to get the final answer, f'(x) = 2xe^(2x) + 2e^(2x) + 2xe^(2x) = 4xe^(2x) + 2e^(2x) Thus, the derivative of the specified function is 4xe^(2x) + 2e^(2x).</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 dividing the function into two parts.</p>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</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 is analyzing the growth of its revenue using the function R(x) = 2xe^x, where x represents time in months. Calculate the rate of change of revenue at x = 3 months.</p>
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<p>A company is analyzing the growth of its revenue using the function R(x) = 2xe^x, where x represents time in months. Calculate the rate of change of revenue at x = 3 months.</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 (revenue function)...(1) Now, we will differentiate the equation (1) Take the derivative of 2xe^x: dR/dx = 2xe^x + 2e^x Given x = 3 (substitute this into the derivative) dR/dx = 2(3)e^3 + 2e^3 = 6e^3 + 2e^3 = 8e^3 Hence, the rate of change of revenue at x = 3 months is 8e^3.</p>
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<p>We have R(x) = 2xe^x (revenue function)...(1) Now, we will differentiate the equation (1) Take the derivative of 2xe^x: dR/dx = 2xe^x + 2e^x Given x = 3 (substitute this into the derivative) dR/dx = 2(3)e^3 + 2e^3 = 6e^3 + 2e^3 = 8e^3 Hence, the rate of change of revenue at x = 3 months is 8e^3.</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 = 3 months as 8e3, indicating that at the specified time, the revenue increases significantly due to the exponential growth factor.</p>
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<p>We find the rate of change of revenue at x = 3 months as 8e3, indicating that at the specified time, the revenue increases significantly due to the exponential growth factor.</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 R(x) = 2xe^x.</p>
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<p>Derive the second derivative of the function R(x) = 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, dR/dx = 2xe^x + 2e^x...(1) Now we will differentiate equation (1) to get the second derivative: d²R/dx² = d/dx [2xe^x + 2e^x] = d/dx [2xe^x] + d/dx [2e^x] Using the product rule on 2xe^x: = 2xe^x + 2e^x + 2e^x = 2xe^x + 4e^x Therefore, the second derivative of the function R(x) = 2xe^x is 2xe^x + 4e^x.</p>
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<p>The first step is to find the first derivative, dR/dx = 2xe^x + 2e^x...(1) Now we will differentiate equation (1) to get the second derivative: d²R/dx² = d/dx [2xe^x + 2e^x] = d/dx [2xe^x] + d/dx [2e^x] Using the product rule on 2xe^x: = 2xe^x + 2e^x + 2e^x = 2xe^x + 4e^x Therefore, the second derivative of the function R(x) = 2xe^x is 2xe^x + 4e^x.</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.</p>
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<p>We use the step-by-step process, where we start with the first derivative.</p>
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<p>Using differentiation rules, we simplify the terms to find the second derivative.</p>
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<p>Using differentiation rules, we simplify the terms 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 ((2xe^x)^2) = 8xe^x(2xe^x + e^x).</p>
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<p>Prove: d/dx ((2xe^x)^2) = 8xe^x(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>Let’s start using the chain rule: Consider y = (2xe^x)^2 To differentiate, we use the chain rule: dy/dx = 2(2xe^x)·d/dx [2xe^x] Since the derivative of 2xe^x is 2xe^x + 2e^x, dy/dx = 2(2xe^x)(2xe^x + 2e^x) = 8xe^x(2xe^x + e^x) Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = (2xe^x)^2 To differentiate, we use the chain rule: dy/dx = 2(2xe^x)·d/dx [2xe^x] Since the derivative of 2xe^x is 2xe^x + 2e^x, dy/dx = 2(2xe^x)(2xe^x + 2e^x) = 8xe^x(2xe^x + e^x) 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 chain rule to differentiate the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation.</p>
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<p>Then, we replace 2xex with its derivative.</p>
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<p>Then, we replace 2xex with its derivative.</p>
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<p>As a final step, we simplify to derive the equation.</p>
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<p>As a final step, we simplify to derive the equation.</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 We will substitute d/dx (2xe^x) = 2xe^x + 2e^x and d/dx (x) = 1 = [(2xe^x + 2e^x)·x - 2xe^x] / x² = [2xe^x·x + 2e^x·x - 2xe^x]/ x² = [2x²e^x + 2xe^x - 2xe^x]/ x² = 2x²e^x/ x² = 2e^x Therefore, d/dx (2xe^x/x) = 2e^x</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 We will substitute d/dx (2xe^x) = 2xe^x + 2e^x and d/dx (x) = 1 = [(2xe^x + 2e^x)·x - 2xe^x] / x² = [2xe^x·x + 2e^x·x - 2xe^x]/ x² = [2x²e^x + 2xe^x - 2xe^x]/ x² = 2x²e^x/ x² = 2e^x 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.</p>
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<p>In this process, we differentiate the given function using the product rule and quotient rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>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 on 2xe^x, we get: d/dx (2xe^x) = 2xe^x + 2e^x</p>
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<p>Using the product rule on 2xe^x, we get: d/dx (2xe^x) = 2xe^x + 2e^x</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, we can use the derivative of 2xe^x in real-life applications, such as analyzing growth rates and changes in systems that follow exponential trends.</p>
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<p>Yes, we can use the derivative of 2xe^x in real-life applications, such as analyzing growth rates and changes in systems that follow exponential trends.</p>
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<h3>3.Is it possible to take the derivative of 2xe^x at x = 0?</h3>
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<h3>3.Is it possible to take the derivative of 2xe^x at x = 0?</h3>
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<p>Yes, at x = 0, the derivative of 2xe^x is 2, as the term involving x becomes zero and only the constant term remains.</p>
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<p>Yes, at x = 0, the derivative of 2xe^x is 2, as the term involving x becomes zero and only the constant term remains.</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 quotient rule to differentiate 2xe^x/x: d/dx (2xe^x/x) = [x(2xe^x + 2e^x) - 2xe^x]/ x²</p>
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<p>We use the quotient rule to differentiate 2xe^x/x: d/dx (2xe^x/x) = [x(2xe^x + 2e^x) - 2xe^x]/ x²</p>
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<h3>5.Are the derivatives of 2xe^x and (e^x)^2 the same?</h3>
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<h3>5.Are the derivatives of 2xe^x and (e^x)^2 the same?</h3>
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<p>No, they are different. The derivative of 2xe^x is 2xe^x + 2e^x, while the derivative of (e^x)^2 is 2e^x(e^x) = 2e^(2x).</p>
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<p>No, they are different. The derivative of 2xe^x is 2xe^x + 2e^x, while the derivative of (e^x)^2 is 2e^x(e^x) = 2e^(2x).</p>
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<h3>6.Can we find the derivative of the 2xe^x formula?</h3>
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<h3>6.Can we find the derivative of the 2xe^x formula?</h3>
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<p>To find, consider f(x) = 2xe^x. We use the product rule: f'(x) = d/dx (2x)·e^x + 2x·d/dx (e^x) = 2e^x + 2xe^x = 2xe^x + 2e^x</p>
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<p>To find, consider f(x) = 2xe^x. We use the product rule: f'(x) = d/dx (2x)·e^x + 2x·d/dx (e^x) = 2e^x + 2xe^x = 2xe^x + 2e^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 mathematical function involving exponents, often represented as ex in natural logarithms.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A mathematical function involving exponents, often represented as ex in natural logarithms.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate functions that are products of two or more functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate functions that are products of two or more functions.</li>
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</ul><ul><li><strong>First Principle:</strong>A method to derive the derivative using the limit of the difference quotient.</li>
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</ul><ul><li><strong>First Principle:</strong>A method to derive the derivative using the limit of the difference quotient.</li>
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</ul><ul><li><strong>Rate of Change:</strong>A measure of how a quantity changes over time or in relation to another variable.</li>
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</ul><ul><li><strong>Rate of Change:</strong>A measure of how a quantity changes over time or in relation to another variable.</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>