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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^(x²) is used to determine how the function e^(x²) changes in response to a slight change in x. Derivatives are crucial in calculating various real-life situations, like rates of change in physics or cost functions in economics. We will now discuss the derivative of e^(x²) in detail.</p>
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<p>The derivative of e^(x²) is used to determine how the function e^(x²) changes in response to a slight change in x. Derivatives are crucial in calculating various real-life situations, like rates of change in physics or cost functions in economics. We will now discuss the derivative of e^(x²) in detail.</p>
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<h2>What is the Derivative of e^x²?</h2>
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<h2>What is the Derivative of e^x²?</h2>
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<p>We aim to understand the derivative of e^(x²). It is commonly represented as d/dx (e^(x²)) or (e^(x²))', and its value is 2x·e^(x²).</p>
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<p>We aim to understand the derivative of e^(x²). It is commonly represented as d/dx (e^(x²)) or (e^(x²))', and its value is 2x·e^(x²).</p>
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<p>The<a>function</a>e^(x²) has a clear derivative, indicating that it is differentiable over its entire domain. Key concepts include:</p>
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<p>The<a>function</a>e^(x²) has a clear derivative, indicating that it is differentiable over its entire domain. Key concepts include:</p>
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<p><strong>Exponential Function:</strong>e^(x²), where e is the<a>base</a>of natural<a>logarithms</a>.</p>
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<p><strong>Exponential Function:</strong>e^(x²), where e is the<a>base</a>of natural<a>logarithms</a>.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating composite functions, essential for e^(x²).</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating composite functions, essential for e^(x²).</p>
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<p><strong>Product Rule:</strong>Used in some methods for differentiation, though not directly for e^(x²).</p>
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<p><strong>Product Rule:</strong>Used in some methods for differentiation, though not directly for e^(x²).</p>
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<h2>Derivative of e^x² Formula</h2>
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<h2>Derivative of e^x² Formula</h2>
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<p>The derivative of e^(x²) is denoted as d/dx (e^(x²)) or (e^(x²))'.</p>
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<p>The derivative of e^(x²) is denoted as d/dx (e^(x²)) or (e^(x²))'.</p>
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<p>The<a>formula</a>used to differentiate e^(x²) is: d/dx (e^(x²)) = 2x·e^(x²)</p>
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<p>The<a>formula</a>used to differentiate e^(x²) is: d/dx (e^(x²)) = 2x·e^(x²)</p>
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<p>This formula applies to all x in the<a>real number</a>domain, as e^(x²) is defined everywhere.</p>
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<p>This formula applies to all x in the<a>real number</a>domain, as e^(x²) is defined everywhere.</p>
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<h2>Proofs of the Derivative of e^x²</h2>
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<h2>Proofs of the Derivative of e^x²</h2>
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<p>The derivative of e^(x²) can be derived using different methods. Here, we'll use the chain rule and the first principle to demonstrate it:</p>
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<p>The derivative of e^(x²) can be derived using different methods. Here, we'll use the chain rule and the first principle to demonstrate it:</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^(x²) using the chain rule: Let u = x², then the function becomes e^u. The derivative of e^u with respect to u is e^u. The derivative of u = x² with respect to x is 2x.</p>
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<p>To prove the differentiation of e^(x²) using the chain rule: Let u = x², then the function becomes e^u. The derivative of e^u with respect to u is e^u. The derivative of u = x² with respect to x is 2x.</p>
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<p>By the chain rule, d/dx (e^(x²)) = e^u · du/dx = e^(x²) · 2x = 2x·e^(x²).</p>
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<p>By the chain rule, d/dx (e^(x²)) = e^u · du/dx = e^(x²) · 2x = 2x·e^(x²).</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>We can also derive the derivative using the first principle, which expresses the derivative as the limit of the difference<a>quotient</a>. Let f(x) = e^(x²).</p>
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<p>We can also derive the derivative using the first principle, which expresses the derivative as the limit of the difference<a>quotient</a>. Let f(x) = e^(x²).</p>
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<p>Its derivative can be expressed as: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [e^((x+h)²) - e^(x²)] / h = limₕ→₀ [e^(x² + 2xh + h²) - e^(x²)] / h = limₕ→₀ [e^(x²) · (e^(2xh + h²) - 1)] / h</p>
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<p>Its derivative can be expressed as: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [e^((x+h)²) - e^(x²)] / h = limₕ→₀ [e^(x² + 2xh + h²) - e^(x²)] / h = limₕ→₀ [e^(x²) · (e^(2xh + h²) - 1)] / h</p>
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<p>Using the expansion of e^u for small u, e^u ≈ 1 + u, we have: = limₕ→₀ [e^(x²) · (2xh + h²)] / h = limₕ→₀ [e^(x²) · (2x + h)] = e^(x²) · 2x</p>
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<p>Using the expansion of e^u for small u, e^u ≈ 1 + u, we have: = limₕ→₀ [e^(x²) · (2xh + h²)] / h = limₕ→₀ [e^(x²) · (2x + h)] = e^(x²) · 2x</p>
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<p>Hence, the derivative of e^(x²) is 2x·e^(x²).</p>
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<p>Hence, the derivative of e^(x²) is 2x·e^(x²).</p>
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<h2>Higher-Order Derivatives of e^x²</h2>
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<h2>Higher-Order Derivatives of e^x²</h2>
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<p>When a function is differentiated<a>multiple</a>times, the subsequent derivatives are called higher-order derivatives. Understanding these derivatives provides deeper insights into the function's behavior.</p>
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<p>When a function is differentiated<a>multiple</a>times, the subsequent derivatives are called higher-order derivatives. Understanding these derivatives provides deeper insights into the function's behavior.</p>
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<p>For the first derivative of a function, we denote it as f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x), which shows the<a>rate</a>of change of the rate of change. Similarly, the third derivative, f′′′(x), results from the second derivative, and this pattern continues.</p>
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<p>For the first derivative of a function, we denote it as f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x), which shows the<a>rate</a>of change of the rate of change. Similarly, the third derivative, f′′′(x), results from the second derivative, and this pattern continues.</p>
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<p>For the nth derivative of e^(x²), we generally use fⁿ(x) to denote the nth derivative of a function f(x), which tells us about the change in the rate of change, continuing for higher-order derivatives.</p>
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<p>For the nth derivative of e^(x²), we generally use fⁿ(x) to denote the nth derivative of a function f(x), which tells us about 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>Since e^(x²) is defined for all real x, there are no points where the derivative is undefined.</p>
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<p>Since e^(x²) is defined for all real x, there are no points where the derivative is undefined.</p>
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<p>At x = 0, the derivative of e^(x²) = 2x·e^(x²) simplifies to 0·e^(0) = 0, indicating no change at that point.</p>
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<p>At x = 0, the derivative of e^(x²) = 2x·e^(x²) simplifies to 0·e^(0) = 0, indicating no change at that point.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^x²</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^x²</h2>
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<p>Students frequently make mistakes when differentiating e^(x²). 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 e^(x²). 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 e^(x²)cos(x).</p>
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<p>Calculate the derivative of e^(x²)cos(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) = e^(x²)cos(x).</p>
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<p>Here, we have f(x) = e^(x²)cos(x).</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^(x²) and v = cos(x).</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^(x²) and v = cos(x).</p>
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<p>Let’s differentiate each term, u′ = d/dx (e^(x²)) = 2x·e^(x²) v′ = d/dx (cos(x)) = -sin(x)</p>
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<p>Let’s differentiate each term, u′ = d/dx (e^(x²)) = 2x·e^(x²) v′ = d/dx (cos(x)) = -sin(x)</p>
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<p>Substituting into the given equation, f'(x) = (2x·e^(x²))(cos(x)) + (e^(x²))(-sin(x))</p>
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<p>Substituting into the given equation, f'(x) = (2x·e^(x²))(cos(x)) + (e^(x²))(-sin(x))</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2x·e^(x²)cos(x) - e^(x²)sin(x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2x·e^(x²)cos(x) - e^(x²)sin(x)</p>
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<p>Thus, the derivative of the specified function is 2x·e^(x²)cos(x) - e^(x²)sin(x).</p>
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<p>Thus, the derivative of the specified function is 2x·e^(x²)cos(x) - e^(x²)sin(x).</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. 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>We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.</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's revenue is modeled by the function R(x) = e^(x²)x. Calculate the rate of change of revenue when x = 1.</p>
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<p>A company's revenue is modeled by the function R(x) = e^(x²)x. Calculate the rate of change of revenue when x = 1.</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) = e^(x²)x (revenue function)...(1)</p>
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<p>We have R(x) = e^(x²)x (revenue function)...(1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Using the product rule, R'(x) = d/dx (e^(x²)x) = (d/dx (e^(x²)))x + e^(x²)(d/dx (x)) = (2x·e^(x²))x + e^(x²)(1) = 2x²·e^(x²) + e^(x²)</p>
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<p>Using the product rule, R'(x) = d/dx (e^(x²)x) = (d/dx (e^(x²)))x + e^(x²)(d/dx (x)) = (2x·e^(x²))x + e^(x²)(1) = 2x²·e^(x²) + e^(x²)</p>
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<p>Given x = 1, substitute this into the derivative: R'(1) = 2(1)²·e^(1²) + e^(1²) = 2·e + e = 3e</p>
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<p>Given x = 1, substitute this into the derivative: R'(1) = 2(1)²·e^(1²) + e^(1²) = 2·e + e = 3e</p>
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<p>Thus, the rate of change of revenue at x = 1 is 3e.</p>
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<p>Thus, the rate of change of revenue at x = 1 is 3e.</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 by differentiating the given revenue function using the product rule. We then substitute x = 1 into the derivative to get the final result.</p>
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<p>We find the rate of change of revenue by differentiating the given revenue function using the product rule. We then substitute x = 1 into the derivative 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 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = e^(x²).</p>
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<p>Derive the second derivative of the function y = e^(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>The first step is to find the first derivative, dy/dx = 2x·e^(x²)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 2x·e^(x²)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2x·e^(x²)]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2x·e^(x²)]</p>
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<p>Here we use the product rule, d²y/dx² = 2(e^(x²) + x(2x·e^(x²))) = 2e^(x²) + 4x²·e^(x²)</p>
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<p>Here we use the product rule, d²y/dx² = 2(e^(x²) + x(2x·e^(x²))) = 2e^(x²) + 4x²·e^(x²)</p>
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<p>Therefore, the second derivative of the function y = e^(x²) is 2e^(x²) + 4x²·e^(x²).</p>
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<p>Therefore, the second derivative of the function y = e^(x²) is 2e^(x²) + 4x²·e^(x²).</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, 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, 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 (e^(x²)tan(x)) = 2x·e^(x²)tan(x) + e^(x²)sec²(x).</p>
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<p>Prove: d/dx (e^(x²)tan(x)) = 2x·e^(x²)tan(x) + e^(x²)sec²(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 product rule: Consider y = e^(x²)tan(x)</p>
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<p>Let’s start using the product rule: Consider y = e^(x²)tan(x)</p>
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<p>To differentiate, we use the product rule: dy/dx = d/dx (e^(x²))tan(x) + e^(x²)d/dx (tan(x)) = (2x·e^(x²))tan(x) + e^(x²)sec²(x)</p>
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<p>To differentiate, we use the product rule: dy/dx = d/dx (e^(x²))tan(x) + e^(x²)d/dx (tan(x)) = (2x·e^(x²))tan(x) + e^(x²)sec²(x)</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 use the product rule to differentiate the equation. We replace each term with its derivative and simplify to derive the equation.</p>
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<p>In this step-by-step process, we use the product rule to differentiate the equation. We replace each term with its derivative and 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 (e^(x²)/x).</p>
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<p>Solve: d/dx (e^(x²)/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 (e^(x²)/x) = (d/dx (e^(x²))x - e^(x²)d/dx (x)) / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (e^(x²)/x) = (d/dx (e^(x²))x - e^(x²)d/dx (x)) / x²</p>
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<p>We will substitute d/dx (e^(x²)) = 2x·e^(x²) and d/dx (x) = 1: = (2x·e^(x²)x - e^(x²)·1) / x² = (2x²·e^(x²) - e^(x²)) / x² = 2x·e^(x²) - e^(x²)/x²</p>
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<p>We will substitute d/dx (e^(x²)) = 2x·e^(x²) and d/dx (x) = 1: = (2x·e^(x²)x - e^(x²)·1) / x² = (2x²·e^(x²) - e^(x²)) / x² = 2x·e^(x²) - e^(x²)/x²</p>
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<p>Therefore, d/dx (e^(x²)/x) = (2x²·e^(x²) - e^(x²)) / x².</p>
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<p>Therefore, d/dx (e^(x²)/x) = (2x²·e^(x²) - e^(x²)) / x².</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 quotient rule. We then simplify the equation to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the quotient rule. We then 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 e^x²</h2>
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<h2>FAQs on the Derivative of e^x²</h2>
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<h3>1.Find the derivative of e^(x²).</h3>
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<h3>1.Find the derivative of e^(x²).</h3>
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<p>Using the chain rule for e^(x²), we get: d/dx (e^(x²)) = 2x·e^(x²).</p>
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<p>Using the chain rule for e^(x²), we get: d/dx (e^(x²)) = 2x·e^(x²).</p>
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<h3>2.Can we use the derivative of e^(x²) in real life?</h3>
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<h3>2.Can we use the derivative of e^(x²) in real life?</h3>
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<p>Yes, the derivative of e^(x²) can be used in real-life scenarios such as modeling population growth, physics problems involving acceleration, and various other applications.</p>
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<p>Yes, the derivative of e^(x²) can be used in real-life scenarios such as modeling population growth, physics problems involving acceleration, and various other applications.</p>
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<h3>3.Is it possible to take the derivative of e^(x²) at any point?</h3>
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<h3>3.Is it possible to take the derivative of e^(x²) at any point?</h3>
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<p>Yes, e^(x²) is defined for all real x, so its derivative can be calculated at any point.</p>
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<p>Yes, e^(x²) is defined for all real x, so its derivative can be calculated at any point.</p>
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<h3>4.What rule is used to differentiate e^(x²)/x?</h3>
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<h3>4.What rule is used to differentiate e^(x²)/x?</h3>
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<p>We use the quotient rule to differentiate e^(x²)/x: d/dx (e^(x²)/x) = (x·(2x·e^(x²)) - e^(x²)·1) / x².</p>
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<p>We use the quotient rule to differentiate e^(x²)/x: d/dx (e^(x²)/x) = (x·(2x·e^(x²)) - e^(x²)·1) / x².</p>
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<h3>5.Are the derivatives of e^(x²) and e^(x)² the same?</h3>
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<h3>5.Are the derivatives of e^(x²) and e^(x)² the same?</h3>
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<p>No, they are different. The derivative of e^(x²) is 2x·e^(x²), while the derivative of e^(x)² requires using the chain rule for<a>powers</a>, yielding 2e^(2x)x.</p>
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<p>No, they are different. The derivative of e^(x²) is 2x·e^(x²), while the derivative of e^(x)² requires using the chain rule for<a>powers</a>, yielding 2e^(2x)x.</p>
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<h2>Important Glossaries for the Derivative of e^x²</h2>
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<h2>Important Glossaries for the Derivative of e^x²</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the 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 function changes in response to a slight change in x.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function in which an independent variable appears in the exponent, such as e^(x²).</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function in which an independent variable appears in the exponent, such as e^(x²).</li>
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</ul><ul><li><strong>Chain Rule:</strong>A differentiation rule used for computing the derivative of the composition of two or more functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A differentiation rule used for computing the derivative of the composition of two or more functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A differentiation rule used when differentiating the product of two functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A differentiation rule used when differentiating the product of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A differentiation rule used when differentiating the quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A differentiation rule used when differentiating the quotient of two functions.</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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<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>