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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>We use the derivative of e^-x^3 as a measuring tool for how the 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 e^-x^3 in detail.</p>
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<p>We use the derivative of e^-x^3 as a measuring tool for how the 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 e^-x^3 in detail.</p>
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<h2>What is the Derivative of e^-x^3?</h2>
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<h2>What is the Derivative of e^-x^3?</h2>
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<p>We now understand the derivative of e^-x^3. It is commonly represented as d/dx (e^-x^3) or (e^-x^3)', and its value is -3x²e^-x^3. The<a>function</a>e^-x^3 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>We now understand the derivative of e^-x^3. It is commonly represented as d/dx (e^-x^3) or (e^-x^3)', and its value is -3x²e^-x^3. The<a>function</a>e^-x^3 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p><strong>Exponential Function:</strong>(e^u, where u = -x^3).</p>
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<p><strong>Exponential Function:</strong>(e^u, where u = -x^3).</p>
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<p><strong>Chain Rule:</strong>Rule for differentiating e^-x^3 since it involves a composite function.</p>
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<p><strong>Chain Rule:</strong>Rule for differentiating e^-x^3 since it involves a composite function.</p>
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<p><strong>Power Rule:</strong>Used in conjunction with the chain rule to differentiate -x^3.</p>
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<p><strong>Power Rule:</strong>Used in conjunction with the chain rule to differentiate -x^3.</p>
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<h2>Derivative of e^-x^3 Formula</h2>
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<h2>Derivative of e^-x^3 Formula</h2>
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<p>The derivative of e^-x^3 can be denoted as d/dx (e^-x^3) or (e^-x^3)'.</p>
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<p>The derivative of e^-x^3 can be denoted as d/dx (e^-x^3) or (e^-x^3)'.</p>
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<p>The<a>formula</a>we use to differentiate e^-x^3 is: d/dx (e^-x^3) = -3x²e^-x^3</p>
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<p>The<a>formula</a>we use to differentiate e^-x^3 is: d/dx (e^-x^3) = -3x²e^-x^3</p>
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<p>The formula applies to all real x.</p>
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<p>The formula applies to all real x.</p>
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<h2>Proofs of the Derivative of e^-x^3</h2>
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<h2>Proofs of the Derivative of e^-x^3</h2>
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<p>We can derive the derivative of e^-x^3 using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of e^-x^3 using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:</p>
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<ol><li>Using Chain Rule</li>
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<ol><li>Using Chain Rule</li>
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<li>Using Product Rule</li>
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<li>Using Product Rule</li>
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</ol><p>We will demonstrate that the differentiation of e^-x^3 results in -3x²e^-x^3 using these methods:</p>
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</ol><p>We will demonstrate that the differentiation of e^-x^3 results in -3x²e^-x^3 using these methods:</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^-x^3 using the chain rule, Consider f(x) = e^u where u = -x^3. Using the chain rule, d/dx [e^u] = e^u · du/dx.</p>
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<p>To prove the differentiation of e^-x^3 using the chain rule, Consider f(x) = e^u where u = -x^3. Using the chain rule, d/dx [e^u] = e^u · du/dx.</p>
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<p>Let's differentiate u = -x^3: du/dx = -3x².</p>
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<p>Let's differentiate u = -x^3: du/dx = -3x².</p>
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<p>Substituting back, d/dx (e^-x^3) = e^-x^3 · (-3x²) = -3x²e^-x^3.</p>
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<p>Substituting back, d/dx (e^-x^3) = e^-x^3 · (-3x²) = -3x²e^-x^3.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Product Rule</h3>
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<h3>Using Product Rule</h3>
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<p>The<a>product</a>rule isn't directly applicable to e^-x^3 as it is a single function, not a product of two functions, so we rely on the chain rule here.</p>
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<p>The<a>product</a>rule isn't directly applicable to e^-x^3 as it is a single function, not a product of two functions, so we rely on the chain rule here.</p>
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<h2>Higher-Order Derivatives of e^-x^3</h2>
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<h2>Higher-Order Derivatives of e^-x^3</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 e^-x^3.</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 e^-x^3.</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 e^-x^3, 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 e^-x^3, 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 e^-x^3 is -3(0)²e^0, which is 0. For large positive or negative x, the exponential<a>term</a>e^-x^3 rapidly approaches 0, making the derivative approach 0 as well.</p>
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<p>When x = 0, the derivative of e^-x^3 is -3(0)²e^0, which is 0. For large positive or negative x, the exponential<a>term</a>e^-x^3 rapidly approaches 0, making the derivative approach 0 as well.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^-x^3</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^-x^3</h2>
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<p>Students frequently make mistakes when differentiating e^-x^3. 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^-x^3. 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^-x^3 · x^2)</p>
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<p>Calculate the derivative of (e^-x^3 · x^2)</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^-x^3 · x².</p>
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<p>Here, we have f(x) = e^-x^3 · x².</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-x^3 and v = x².</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-x^3 and v = x².</p>
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<p>Let's differentiate each term, u′ = d/dx (e^-x^3) = -3x²e^-x^3 v′ = d/dx (x²) = 2x</p>
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<p>Let's differentiate each term, u′ = d/dx (e^-x^3) = -3x²e^-x^3 v′ = d/dx (x²) = 2x</p>
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<p>Substituting into the given equation, f'(x) = (-3x²e^-x^3) · x² + e^-x^3 · 2x</p>
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<p>Substituting into the given equation, f'(x) = (-3x²e^-x^3) · x² + e^-x^3 · 2x</p>
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<p>Let's simplify terms to get the final answer, f'(x) = -3x^4e^-x^3 + 2xe^-x^3</p>
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<p>Let's simplify terms to get the final answer, f'(x) = -3x^4e^-x^3 + 2xe^-x^3</p>
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<p>Thus, the derivative of the specified function is -3x^4e^-x^3 + 2xe^-x^3.</p>
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<p>Thus, the derivative of the specified function is -3x^4e^-x^3 + 2xe^-x^3.</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>An engineering company is modeling the cooling of a material using the function T(x) = e^-x^3 where T represents the temperature. If x = 1, find the rate of change of temperature.</p>
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<p>An engineering company is modeling the cooling of a material using the function T(x) = e^-x^3 where T represents the temperature. If x = 1, find the rate of change of temperature.</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 T(x) = e^-x^3 (temperature model)...(1)</p>
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<p>We have T(x) = e^-x^3 (temperature model)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of e^-x^3: dT/dx = -3x²e^-x^3</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of e^-x^3: dT/dx = -3x²e^-x^3</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>dT/dx = -3(1)²e^-1^3 = -3e^-1</p>
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<p>dT/dx = -3(1)²e^-1^3 = -3e^-1</p>
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<p>Hence, the rate of change of temperature at x = 1 is -3/e.</p>
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<p>Hence, the rate of change of temperature at x = 1 is -3/e.</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 temperature at x = 1 as -3/e, which means that the temperature decreases at this rate as x increases.</p>
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<p>We find the rate of change of temperature at x = 1 as -3/e, which means that the temperature decreases at this rate as x increases.</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 f(x) = e^-x^3.</p>
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<p>Derive the second derivative of the function f(x) = e^-x^3.</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, f'(x) = -3x²e^-x^3...(1)</p>
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<p>The first step is to find the first derivative, f'(x) = -3x²e^-x^3...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: f''(x) = d/dx [-3x²e^-x^3]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: f''(x) = d/dx [-3x²e^-x^3]</p>
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<p>Using the product rule, f''(x) = d/dx [-3x²] · e^-x^3 + (-3x²) · d/dx [e^-x^3] = [-6x · e^-x^3] + (-3x²) · [-3x²e^-x^3] = -6xe^-x^3 + 9x^4e^-x^3</p>
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<p>Using the product rule, f''(x) = d/dx [-3x²] · e^-x^3 + (-3x²) · d/dx [e^-x^3] = [-6x · e^-x^3] + (-3x²) · [-3x²e^-x^3] = -6xe^-x^3 + 9x^4e^-x^3</p>
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<p>Therefore, the second derivative of the function f(x) = e^-x^3 is -6xe^-x^3 + 9x^4e^-x^3.</p>
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<p>Therefore, the second derivative of the function f(x) = e^-x^3 is -6xe^-x^3 + 9x^4e^-x^3.</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 -3x² and e^-x^3 separately. We then substitute and simplify the terms to find the final answer.</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 -3x² and e^-x^3 separately. We then substitute and simplify the terms to find the final answer.</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 (x^3e^-x^3) = (3x² - x^5)e^-x^3.</p>
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<p>Prove: d/dx (x^3e^-x^3) = (3x² - x^5)e^-x^3.</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 = x^3e^-x^3.</p>
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<p>Let's start using the product rule: Consider y = x^3e^-x^3.</p>
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<p>To differentiate, we use the product rule: dy/dx = d/dx [x^3] · e^-x^3 + x^3 · d/dx [e^-x^3] = 3x²e^-x^3 + x^3 · (-3x²e^-x^3) = 3x²e^-x^3 - 3x^5e^-x^3 = (3x² - x^5)e^-x^3</p>
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<p>To differentiate, we use the product rule: dy/dx = d/dx [x^3] · e^-x^3 + x^3 · d/dx [e^-x^3] = 3x²e^-x^3 + x^3 · (-3x²e^-x^3) = 3x²e^-x^3 - 3x^5e^-x^3 = (3x² - x^5)e^-x^3</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. We differentiate each term separately and simplify to derive the equation.</p>
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<p>In this step-by-step process, we used the product rule to differentiate the equation. We differentiate each term separately 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^-x^3/x).</p>
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<p>Solve: d/dx (e^-x^3/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^-x^3/x) = (d/dx (e^-x^3) · x - e^-x^3 · d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (e^-x^3/x) = (d/dx (e^-x^3) · x - e^-x^3 · d/dx(x))/x²</p>
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<p>We will substitute d/dx (e^-x^3) = -3x²e^-x^3 and d/dx (x) = 1 = (-3x²e^-x^3 · x - e^-x^3 · 1) / x² = (-3x³e^-x^3 - e^-x^3) / x² = -e^-x^3(3x³ + 1)/x²</p>
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<p>We will substitute d/dx (e^-x^3) = -3x²e^-x^3 and d/dx (x) = 1 = (-3x²e^-x^3 · x - e^-x^3 · 1) / x² = (-3x³e^-x^3 - e^-x^3) / x² = -e^-x^3(3x³ + 1)/x²</p>
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<p>Therefore, d/dx (e^-x^3/x) = -e^-x^3(3x³ + 1)/x².</p>
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<p>Therefore, d/dx (e^-x^3/x) = -e^-x^3(3x³ + 1)/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. 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 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 e^-x^3</h2>
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<h2>FAQs on the Derivative of e^-x^3</h2>
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<h3>1.Find the derivative of e^-x^3.</h3>
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<h3>1.Find the derivative of e^-x^3.</h3>
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<p>Using the chain rule for e^-x^3 gives -3x²e^-x^3.</p>
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<p>Using the chain rule for e^-x^3 gives -3x²e^-x^3.</p>
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<h3>2.Can we use the derivative of e^-x^3 in real life?</h3>
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<h3>2.Can we use the derivative of e^-x^3 in real life?</h3>
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<p>Yes, we can use the derivative of e^-x^3 in real life to model processes like cooling, decay, and other exponential changes in fields such as physics and engineering.</p>
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<p>Yes, we can use the derivative of e^-x^3 in real life to model processes like cooling, decay, and other exponential changes in fields such as physics and engineering.</p>
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<h3>3.Is it possible to take the derivative of e^-x^3 at x = 0?</h3>
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<h3>3.Is it possible to take the derivative of e^-x^3 at x = 0?</h3>
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<p>Yes, it is possible to take the derivative of e^-x^3 at x = 0, and it results in 0 since the derivative is -3x²e^-x^3, which is 0 when x = 0.</p>
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<p>Yes, it is possible to take the derivative of e^-x^3 at x = 0, and it results in 0 since the derivative is -3x²e^-x^3, which is 0 when x = 0.</p>
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<h3>4.What rule is used to differentiate e^-x^3/x?</h3>
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<h3>4.What rule is used to differentiate e^-x^3/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate e^-x^3/x: d/dx (e^-x^3/x) = (-3x³e^-x^3 - e^-x^3)/x².</p>
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<p>We use the<a>quotient</a>rule to differentiate e^-x^3/x: d/dx (e^-x^3/x) = (-3x³e^-x^3 - e^-x^3)/x².</p>
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<h3>5.Are the derivatives of e^-x^3 and e^x^3 the same?</h3>
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<h3>5.Are the derivatives of e^-x^3 and e^x^3 the same?</h3>
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<p>No, they are different. The derivative of e^-x^3 is -3x²e^-x^3, while the derivative of e^x^3 is 3x²e^x^3.</p>
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<p>No, they are different. The derivative of e^-x^3 is -3x²e^-x^3, while the derivative of e^x^3 is 3x²e^x^3.</p>
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<h2>Important Glossaries for the Derivative of e^-x^3</h2>
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<h2>Important Glossaries for the Derivative of e^-x^3</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^u, where e is the base of natural logarithms.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form e^u, where e is the base of natural logarithms.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to find the derivative of a power function.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to find the derivative of a power function.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A method of finding the derivative of a quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A method of finding the derivative of a 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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<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>