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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>The derivative of e^(2/x) provides insight into how the function changes with respect to slight variations in x. Derivatives are crucial in various real-life applications, such as calculating rates of change and modeling exponential growth. In this article, we delve into the derivative of e^(2/x) in detail.</p>
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<p>The derivative of e^(2/x) provides insight into how the function changes with respect to slight variations in x. Derivatives are crucial in various real-life applications, such as calculating rates of change and modeling exponential growth. In this article, we delve into the derivative of e^(2/x) in detail.</p>
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<h2>What is the Derivative of e^(2/x)?</h2>
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<h2>What is the Derivative of e^(2/x)?</h2>
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<p>To understand the derivative<a>of</a>e^(2/x), we represent it as d/dx (e^(2/x)) or (e^(2/x))'.</p>
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<p>To understand the derivative<a>of</a>e^(2/x), we represent it as d/dx (e^(2/x)) or (e^(2/x))'.</p>
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<p>Using the chain rule, the derivative is found to be -2/x² * e^(2/x). The<a>function</a>e^(2/x) is differentiable across its domain, and its derivative is crucial for analyzing<a>exponential growth</a>and decay processes. Key concepts include: -</p>
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<p>Using the chain rule, the derivative is found to be -2/x² * e^(2/x). The<a>function</a>e^(2/x) is differentiable across its domain, and its derivative is crucial for analyzing<a>exponential growth</a>and decay processes. Key concepts include: -</p>
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<p><strong>Exponential Function:</strong>e^u, where u is a function of x. </p>
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<p><strong>Exponential Function:</strong>e^u, where u is a function of x. </p>
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<p><strong>Chain Rule:</strong>A differentiation technique used for composite functions. </p>
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<p><strong>Chain Rule:</strong>A differentiation technique used for composite functions. </p>
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<p><strong>Power Rule:</strong>Used for differentiating functions of the form x^n.</p>
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<p><strong>Power Rule:</strong>Used for differentiating functions of the form x^n.</p>
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<h2>Derivative of e^(2/x) Formula</h2>
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<h2>Derivative of e^(2/x) Formula</h2>
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<p>The derivative of e^(2/x) can be expressed as d/dx (e^(2/x)) or (e^(2/x))'.</p>
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<p>The derivative of e^(2/x) can be expressed as d/dx (e^(2/x)) or (e^(2/x))'.</p>
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<p>The<a>formula</a>to differentiate e^(2/x) is: d/dx (e^(2/x)) = -2/x² * e^(2/x), This formula is applicable for all x ≠ 0.</p>
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<p>The<a>formula</a>to differentiate e^(2/x) is: d/dx (e^(2/x)) = -2/x² * e^(2/x), This formula is applicable for all x ≠ 0.</p>
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<h2>Proofs of the Derivative of e^(2/x)</h2>
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<h2>Proofs of the Derivative of e^(2/x)</h2>
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<p>We can derive the derivative of e^(2/x) using various methods. Here are some approaches: </p>
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<p>We can derive the derivative of e^(2/x) using various methods. Here are some approaches: </p>
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<ol><li>By Chain Rule </li>
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<ol><li>By Chain Rule </li>
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<li>By Product Rule</li>
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<li>By Product Rule</li>
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</ol><p>Let's demonstrate the differentiation of e^(2/x) using the chain rule:</p>
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</ol><p>Let's demonstrate the differentiation of e^(2/x) using the chain rule:</p>
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<h3>Using Chain Rule</h3>
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<h3>Using Chain Rule</h3>
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<p>To prove the differentiation of e^(2/x) using the chain rule, we<a>set</a>: f(x) = 2/x</p>
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<p>To prove the differentiation of e^(2/x) using the chain rule, we<a>set</a>: f(x) = 2/x</p>
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<p>Then e^(2/x) becomes e^f(x).</p>
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<p>Then e^(2/x) becomes e^f(x).</p>
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<p>The differentiation process involves: g(x) = e^u, where u = 2/x</p>
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<p>The differentiation process involves: g(x) = e^u, where u = 2/x</p>
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<p>Differentiating e^u gives: g'(x) = e^u * u' u' = d/dx (2/x) = -2/x²</p>
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<p>Differentiating e^u gives: g'(x) = e^u * u' u' = d/dx (2/x) = -2/x²</p>
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<p>Thus, the derivative of e^(2/x) is: d/dx (e^(2/x)) = e^(2/x) * (-2/x²) = -2/x² * e^(2/x)</p>
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<p>Thus, the derivative of e^(2/x) is: d/dx (e^(2/x)) = e^(2/x) * (-2/x²) = -2/x² * e^(2/x)</p>
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<h2>Higher-Order Derivatives of e^(2/x)</h2>
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<h2>Higher-Order Derivatives of e^(2/x)</h2>
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<p>Higher-order derivatives are obtained by differentiating the first derivative successively. For example, the second derivative involves differentiating the first derivative of e^(2/x) again. Higher-order derivatives are significant in understanding the behavior and curvature of functions like e^(2/x).</p>
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<p>Higher-order derivatives are obtained by differentiating the first derivative successively. For example, the second derivative involves differentiating the first derivative of e^(2/x) again. Higher-order derivatives are significant in understanding the behavior and curvature of functions like e^(2/x).</p>
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<p>For the first derivative, we write f′(x), which gives the<a>rate</a>of change of the function. The second derivative, f′′(x), is derived from the first derivative and provides insights into the concavity of the function.</p>
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<p>For the first derivative, we write f′(x), which gives the<a>rate</a>of change of the function. The second derivative, f′′(x), is derived from the first derivative and provides insights into the concavity of the function.</p>
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<p>The nth Derivative, denoted as fⁿ(x), describes the change in the rate of change and can be useful in various applications.</p>
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<p>The nth Derivative, denoted as fⁿ(x), describes the change in the rate of change and can be useful in various applications.</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 function e^(2/x) is undefined.</p>
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<p>When x = 0, the function e^(2/x) is undefined.</p>
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<p>Therefore, its derivative cannot be evaluated at this point. For positive or negative values of x where x ≠ 0, the derivative of e^(2/x) is -2/x² * e^(2/x).</p>
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<p>Therefore, its derivative cannot be evaluated at this point. For positive or negative values of x where x ≠ 0, the derivative of e^(2/x) is -2/x² * e^(2/x).</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^(2/x)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^(2/x)</h2>
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<p>Students often make errors when differentiating e^(2/x). Understanding the correct process is key to avoiding these mistakes. Here are some common errors and solutions:</p>
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<p>Students often make errors when differentiating e^(2/x). Understanding the correct process is key to avoiding these mistakes. Here are some common errors and solutions:</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 e^(2/x) * x²</p>
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<p>Calculate the derivative of e^(2/x) * x²</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 f(x) = e^(2/x) * x².</p>
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<p>Let f(x) = e^(2/x) * x².</p>
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<p>Using the product rule, f'(x) = u'v + uv' Here, u = e^(2/x) and v = x².</p>
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<p>Using the product rule, f'(x) = u'v + uv' Here, u = e^(2/x) and v = x².</p>
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<p>Differentiate each term: u' = -2/x² * e^(2/x) v' = 2x</p>
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<p>Differentiate each term: u' = -2/x² * e^(2/x) v' = 2x</p>
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<p>Substitute these into the equation: f'(x) = (-2/x² * e^(2/x)) * x² + (e^(2/x)) * 2x</p>
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<p>Substitute these into the equation: f'(x) = (-2/x² * e^(2/x)) * x² + (e^(2/x)) * 2x</p>
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<p>Simplify the terms: f'(x) = -2e^(2/x) + 2xe^(2/x)</p>
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<p>Simplify the terms: f'(x) = -2e^(2/x) + 2xe^(2/x)</p>
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<p>Thus, the derivative is -2e^(2/x) + 2xe^(2/x).</p>
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<p>Thus, the derivative is -2e^(2/x) + 2xe^(2/x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The derivative is calculated by dividing the function into parts and applying the product rule. Each part is differentiated separately and then combined for the final result.</p>
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<p>The derivative is calculated by dividing the function into parts and applying the product rule. Each part is differentiated separately and then combined for 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 bacteria culture grows at a rate modeled by y = e^(2/x), where y is the population size and x is the time in hours. Find the rate of growth at x = 2 hours.</p>
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<p>A bacteria culture grows at a rate modeled by y = e^(2/x), where y is the population size and x is the time in hours. Find the rate of growth at x = 2 hours.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Given y = e^(2/x), Differentiate y with respect to x: dy/dx = -2/x² * e^(2/x)</p>
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<p>Given y = e^(2/x), Differentiate y with respect to x: dy/dx = -2/x² * e^(2/x)</p>
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<p>Substitute x = 2 into the derivative: dy/dx = -2/(2²) * e^(2/2) = -2/4 * e = -1/2 * e</p>
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<p>Substitute x = 2 into the derivative: dy/dx = -2/(2²) * e^(2/2) = -2/4 * e = -1/2 * e</p>
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<p>The rate of growth at x = 2 hours is -1/2 * e.</p>
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<p>The rate of growth at x = 2 hours is -1/2 * e.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>By substituting x = 2 into the derivative, we determine the rate of growth of the bacteria culture at that point in time.</p>
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<p>By substituting x = 2 into the derivative, we determine the rate of growth of the bacteria culture at that point in 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>Find the second derivative of y = e^(2/x).</p>
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<p>Find the second derivative of y = e^(2/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>First, find the first derivative: dy/dx = -2/x² * e^(2/x)</p>
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<p>First, find the first derivative: dy/dx = -2/x² * e^(2/x)</p>
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<p>Now differentiate again: d²y/dx² = d/dx [-2/x² * e^(2/x)]</p>
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<p>Now differentiate again: d²y/dx² = d/dx [-2/x² * e^(2/x)]</p>
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<p>Using the product rule: = [-2/x² * d/dx (e^(2/x)) + e^(2/x) * d/dx (-2/x²)] = [-2/x² * (-2/x² * e^(2/x)) + e^(2/x) * (4/x³)]</p>
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<p>Using the product rule: = [-2/x² * d/dx (e^(2/x)) + e^(2/x) * d/dx (-2/x²)] = [-2/x² * (-2/x² * e^(2/x)) + e^(2/x) * (4/x³)]</p>
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<p>Simplify: = [4/x⁴ * e^(2/x) + 4/x³ * e^(2/x)] = 4/x⁴ * e^(2/x) * (1 + x)</p>
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<p>Simplify: = [4/x⁴ * e^(2/x) + 4/x³ * e^(2/x)] = 4/x⁴ * e^(2/x) * (1 + x)</p>
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<p>Thus, the second derivative is 4/x⁴ * e^(2/x) * (1 + x).</p>
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<p>Thus, the second derivative is 4/x⁴ * e^(2/x) * (1 + x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The second derivative is found by differentiating the first derivative, applying the product rule again, and simplifying the resulting expression.</p>
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<p>The second derivative is found by differentiating the first derivative, applying the product rule again, and simplifying the resulting expression.</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 (e^(4/x)) = -4/x² * e^(4/x).</p>
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<p>Prove: d/dx (e^(4/x)) = -4/x² * e^(4/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 y = e^(4/x). Using the chain rule: u = 4/x Then, y = e^u.</p>
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<p>Let y = e^(4/x). Using the chain rule: u = 4/x Then, y = e^u.</p>
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<p>The derivative is: dy/dx = e^u * du/dx du/dx = -4/x²</p>
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<p>The derivative is: dy/dx = e^u * du/dx du/dx = -4/x²</p>
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<p>Therefore, dy/dx = e^(4/x) * (-4/x²) dy/dx = -4/x² * e^(4/x) Hence proved.</p>
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<p>Therefore, dy/dx = e^(4/x) * (-4/x²) dy/dx = -4/x² * e^(4/x) Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The proof is accomplished by identifying the inner function and applying the chain rule to differentiate e^(4/x).</p>
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<p>The proof is accomplished by identifying the inner function and applying the chain rule to differentiate e^(4/x).</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 (x * e^(2/x))</p>
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<p>Solve: d/dx (x * e^(2/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>Use the product rule: d/dx (x * e^(2/x)) = (d/dx (x) * e^(2/x) + x * d/dx (e^(2/x)))</p>
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<p>Use the product rule: d/dx (x * e^(2/x)) = (d/dx (x) * e^(2/x) + x * d/dx (e^(2/x)))</p>
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<p>Substitute d/dx (x) = 1 and d/dx (e^(2/x)) = -2/x² * e^(2/x): = (1 * e^(2/x) + x * (-2/x² * e^(2/x))) = e^(2/x) - 2/x * e^(2/x) = e^(2/x) * (1 - 2/x)</p>
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<p>Substitute d/dx (x) = 1 and d/dx (e^(2/x)) = -2/x² * e^(2/x): = (1 * e^(2/x) + x * (-2/x² * e^(2/x))) = e^(2/x) - 2/x * e^(2/x) = e^(2/x) * (1 - 2/x)</p>
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<p>Therefore, the derivative is e^(2/x) * (1 - 2/x).</p>
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<p>Therefore, the derivative is e^(2/x) * (1 - 2/x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The function is differentiated using the product rule, simplifying the expression to achieve the final result.</p>
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<p>The function is differentiated using the product rule, simplifying the expression to achieve 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 e^(2/x)</h2>
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<h2>FAQs on the Derivative of e^(2/x)</h2>
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<h3>1.What is the derivative of e^(2/x)?</h3>
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<h3>1.What is the derivative of e^(2/x)?</h3>
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<p>Using the chain rule, the derivative of e^(2/x) is -2/x² * e^(2/x).</p>
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<p>Using the chain rule, the derivative of e^(2/x) is -2/x² * e^(2/x).</p>
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<h3>2.Can the derivative of e^(2/x) be applied in real life?</h3>
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<h3>2.Can the derivative of e^(2/x) be applied in real life?</h3>
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<p>Yes, it can be used to model exponential growth or decay, which is common in fields like biology, economics, and physics.</p>
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<p>Yes, it can be used to model exponential growth or decay, which is common in fields like biology, economics, and physics.</p>
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<h3>3.Is the derivative of e^(2/x) defined at x = 0?</h3>
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<h3>3.Is the derivative of e^(2/x) defined at x = 0?</h3>
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<p>No, the derivative is undefined at x = 0 because e^(2/x) is undefined at this point.</p>
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<p>No, the derivative is undefined at x = 0 because e^(2/x) is undefined at this point.</p>
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<h3>4.What rule is used to differentiate e^(2/x)?</h3>
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<h3>4.What rule is used to differentiate e^(2/x)?</h3>
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<p>The chain rule is used, as it is a composite function where 2/x is the inner function.</p>
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<p>The chain rule is used, as it is a composite function where 2/x is the inner function.</p>
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<h3>5.Are the derivatives of e^(2/x) and ln(x) the same?</h3>
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<h3>5.Are the derivatives of e^(2/x) and ln(x) the same?</h3>
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<p>No, they are different. The derivative of e^(2/x) is -2/x² * e^(2/x), while the derivative of ln(x) is 1/x.</p>
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<p>No, they are different. The derivative of e^(2/x) is -2/x² * e^(2/x), while the derivative of ln(x) is 1/x.</p>
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<h2>Important Glossaries for the Derivative of e^(2/x)</h2>
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<h2>Important Glossaries for the Derivative of e^(2/x)</h2>
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<ul><li><strong>Derivative:</strong>The rate at which a function changes with respect to a variable.</li>
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<ul><li><strong>Derivative:</strong>The rate at which a function changes with respect to a variable.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form e^u, where u is a variable expression.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form e^u, where u is a variable expression.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A differentiation method for composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A differentiation method for composite functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two functions.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point where a function or its derivative does not exist or is not defined.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point where a function or its derivative does not exist or is not defined.</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>