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<p>Last updated on<strong>September 20, 2025</strong></p>
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<p>Last updated on<strong>September 20, 2025</strong></p>
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<p>The derivative of te^t highlights how the function te^t changes in response to a small change in t. Derivatives are essential tools in understanding dynamic systems, especially in fields such as economics and engineering. We will now discuss the derivative of te^t in detail.</p>
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<p>The derivative of te^t highlights how the function te^t changes in response to a small change in t. Derivatives are essential tools in understanding dynamic systems, especially in fields such as economics and engineering. We will now discuss the derivative of te^t in detail.</p>
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<h2>What is the Derivative of te^t?</h2>
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<h2>What is the Derivative of te^t?</h2>
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<p>We now understand the derivative of te^t. It is commonly represented as d/dt (te^t) or (te^t)', and its value is e^t + te^t.</p>
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<p>We now understand the derivative of te^t. It is commonly represented as d/dt (te^t) or (te^t)', and its value is e^t + te^t.</p>
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<p>The<a>function</a>te^t has a clearly defined derivative, indicating it is differentiable within its domain. Key concepts include: </p>
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<p>The<a>function</a>te^t has a clearly defined derivative, indicating it is differentiable within its domain. Key concepts include: </p>
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<p>Exponential Function: (e^t is the exponential function with<a>base</a>e). </p>
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<p>Exponential Function: (e^t is the exponential function with<a>base</a>e). </p>
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<p>Product Rule: Used for differentiating te^t (since it consists of t multiplied by e^t).</p>
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<p>Product Rule: Used for differentiating te^t (since it consists of t multiplied by e^t).</p>
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<h2>Derivative of te^t Formula</h2>
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<h2>Derivative of te^t Formula</h2>
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<p>The derivative of te^t can be denoted as d/dt (te^t) or (te^t)'. The<a>formula</a>we use to differentiate te^t is: d/dt (te^t) = e^t + te^t This formula applies to all t in the<a>real number</a>domain.</p>
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<p>The derivative of te^t can be denoted as d/dt (te^t) or (te^t)'. The<a>formula</a>we use to differentiate te^t is: d/dt (te^t) = e^t + te^t This formula applies to all t in the<a>real number</a>domain.</p>
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<h2>Proofs of the Derivative of te^t</h2>
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<h2>Proofs of the Derivative of te^t</h2>
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<p>We can derive the derivative of te^t using proofs. To show this, we will use differentiation rules. Several methods can prove this: -</p>
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<p>We can derive the derivative of te^t using proofs. To show this, we will use differentiation rules. Several methods can prove this: -</p>
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<ol><li>By First Principle </li>
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<ol><li>By First Principle </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>Let's demonstrate that differentiating te^t results in e^t + te^t using these methods:</p>
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</ol><p>Let's demonstrate that differentiating te^t results in e^t + te^t using these methods:</p>
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<h3>Using First Principle</h3>
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<h3>Using First Principle</h3>
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<p>The derivative of te^t can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>The derivative of te^t can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of te^t using the first principle, consider f(t) = te^t. Its derivative can be expressed as the following limit: f'(t) = limₕ→₀ [f(t + h) - f(t)] / h</p>
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<p>To find the derivative of te^t using the first principle, consider f(t) = te^t. Its derivative can be expressed as the following limit: f'(t) = limₕ→₀ [f(t + h) - f(t)] / h</p>
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<p>Given f(t) = te^t, we write f(t + h) = (t + h)e^(t + h).</p>
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<p>Given f(t) = te^t, we write f(t + h) = (t + h)e^(t + h).</p>
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<p>Substituting these into the<a>equation</a>, f'(t) = limₕ→₀ [(t + h)e^(t + h) - te^t] / h = limₕ→₀ [te^(t + h) + he^(t + h) - te^t] / h = limₕ→₀ [te^(t + h) - te^t + he^(t + h)] / h = limₕ→₀ [t(e^(t + h) - e^t) + he^(t + h)] / h</p>
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<p>Substituting these into the<a>equation</a>, f'(t) = limₕ→₀ [(t + h)e^(t + h) - te^t] / h = limₕ→₀ [te^(t + h) + he^(t + h) - te^t] / h = limₕ→₀ [te^(t + h) - te^t + he^(t + h)] / h = limₕ→₀ [t(e^(t + h) - e^t) + he^(t + h)] / h</p>
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<p>Using the limit property e^(t + h) = e^t * e^h and the fact limₕ→₀ (e^h - 1)/h = 1, f'(t) = te^t * 1 + e^t = e^t + te^t</p>
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<p>Using the limit property e^(t + h) = e^t * e^h and the fact limₕ→₀ (e^h - 1)/h = 1, f'(t) = te^t * 1 + e^t = e^t + te^t</p>
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<h3>Using Product Rule</h3>
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<h3>Using Product Rule</h3>
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<p>To prove the differentiation of te^t using the<a>product</a>rule, Consider f(t) = t and g(t) = e^t so that te^t = f(t)g(t).</p>
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<p>To prove the differentiation of te^t using the<a>product</a>rule, Consider f(t) = t and g(t) = e^t so that te^t = f(t)g(t).</p>
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<p>By the product rule: d/dt [f(t)g(t)] = f'(t)g(t) + f(t)g'(t) f'(t) = 1 (derivative of t) g'(t) = e^t (derivative of e^t)</p>
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<p>By the product rule: d/dt [f(t)g(t)] = f'(t)g(t) + f(t)g'(t) f'(t) = 1 (derivative of t) g'(t) = e^t (derivative of e^t)</p>
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<p>Substituting into the product rule, d/dt (te^t) = 1(e^t) + t(e^t) = e^t + te^t</p>
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<p>Substituting into the product rule, d/dt (te^t) = 1(e^t) + t(e^t) = e^t + te^t</p>
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<h2>Higher-Order Derivatives of te^t</h2>
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<h2>Higher-Order Derivatives of te^t</h2>
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<p>When a function is differentiated<a>multiple</a>times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives offer deeper insights into the behavior<a>of functions</a>.</p>
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<p>When a function is differentiated<a>multiple</a>times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives offer deeper insights into the behavior<a>of functions</a>.</p>
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<p>For instance, consider a vehicle where the velocity changes (first derivative), and the<a>rate</a>at which the velocity changes (second derivative) also varies. Higher-order derivatives help us understand functions like te^t more clearly.</p>
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<p>For instance, consider a vehicle where the velocity changes (first derivative), and the<a>rate</a>at which the velocity changes (second derivative) also varies. Higher-order derivatives help us understand functions like te^t more clearly.</p>
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<p>For the first derivative of a function, we denote it as f′(t). The second derivative, f′′(t), is obtained from the first derivative. Similarly, the third derivative, f′′′(t), results from the second derivative, continuing this pattern.</p>
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<p>For the first derivative of a function, we denote it as f′(t). The second derivative, f′′(t), is obtained from the first derivative. Similarly, the third derivative, f′′′(t), results from the second derivative, continuing this pattern.</p>
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<p>For the nth Derivative of te^t, we generally use f n(t) for the nth derivative of a function f(t), which tells us the change in the rate of change.</p>
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<p>For the nth Derivative of te^t, we generally use f n(t) for the nth derivative of a function f(t), 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 t is 0, the derivative of te^t = e^0 + 0 * e^0, which is 1. - As t approaches -∞, the derivative approaches 0 since e^t diminishes.</p>
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<p>- When t is 0, the derivative of te^t = e^0 + 0 * e^0, which is 1. - As t approaches -∞, the derivative approaches 0 since e^t diminishes.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of te^t</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of te^t</h2>
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<p>Students frequently make mistakes when differentiating te^t. 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 te^t. 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 (te^t + e^t).</p>
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<p>Calculate the derivative of (te^t + e^t).</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(t) = te^t + e^t.</p>
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<p>Here, we have f(t) = te^t + e^t.</p>
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<p>Using the sum rule and product rule, f'(t) = d/dt(te^t) + d/dt(e^t) = (e^t + te^t) + e^t = e^t + te^t + e^t = 2e^t + te^t</p>
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<p>Using the sum rule and product rule, f'(t) = d/dt(te^t) + d/dt(e^t) = (e^t + te^t) + e^t = e^t + te^t + e^t = 2e^t + te^t</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 applying the sum rule and the product rule. The first step is to differentiate each part and then combine the results to get the final derivative.</p>
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<p>We find the derivative of the given function by applying the sum rule and the product rule. The first step is to differentiate each part and then combine the results to get the final derivative.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A company models its revenue growth over time using the function R(t) = te^t. Calculate the rate of revenue growth when t = 2.</p>
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<p>A company models its revenue growth over time using the function R(t) = te^t. Calculate the rate of revenue growth when t = 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>We have R(t) = te^t (revenue growth function).</p>
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<p>We have R(t) = te^t (revenue growth function).</p>
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<p>Differentiate R(t): dR/dt = e^t + te^t</p>
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<p>Differentiate R(t): dR/dt = e^t + te^t</p>
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<p>Substitute t = 2: dR/dt = e^2 + 2e^2 = 3e^2</p>
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<p>Substitute t = 2: dR/dt = e^2 + 2e^2 = 3e^2</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 revenue growth at t = 2 by differentiating the revenue function and substituting t = 2 into the derivative. This provides the rate at which revenue grows at that point in time.</p>
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<p>We find the rate of revenue growth at t = 2 by differentiating the revenue function and substituting t = 2 into the derivative. This provides the rate at which revenue grows 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>Derive the second derivative of the function f(t) = te^t.</p>
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<p>Derive the second derivative of the function f(t) = te^t.</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'(t) = e^t + te^t</p>
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<p>The first step is to find the first derivative, f'(t) = e^t + te^t</p>
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<p>Now differentiate f'(t) to get the second derivative: f''(t) = d/dt(e^t + te^t) = e^t + (e^t + te^t) = 2e^t + te^t</p>
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<p>Now differentiate f'(t) to get the second derivative: f''(t) = d/dt(e^t + te^t) = e^t + (e^t + te^t) = 2e^t + te^t</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We start with the first derivative of f(t) and then differentiate it again to find the second derivative. Using the product and sum rules, we arrive at the final expression.</p>
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<p>We start with the first derivative of f(t) and then differentiate it again to find the second derivative. Using the product and sum rules, we arrive at the final 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/dt (t^2e^t) = 2te^t + t^2e^t.</p>
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<p>Prove: d/dt (t^2e^t) = 2te^t + t^2e^t.</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 = t^2e^t</p>
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<p>Let’s start using the product rule: Consider y = t^2e^t</p>
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<p>To differentiate, apply the product rule:</p>
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<p>To differentiate, apply the product rule:</p>
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<p>dy/dt = d/dt(t^2) * e^t + t^2 * d/dt(e^t) = 2t * e^t + t^2 * e^t = 2te^t + t^2e^t</p>
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<p>dy/dt = d/dt(t^2) * e^t + t^2 * d/dt(e^t) = 2t * e^t + t^2 * e^t = 2te^t + t^2e^t</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the product rule to differentiate t^2e^t. Each term is differentiated separately, and then the results are combined to prove the required expression.</p>
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<p>We use the product rule to differentiate t^2e^t. Each term is differentiated separately, and then the results are combined to prove the required expression.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dt (te^t/t).</p>
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<p>Solve: d/dt (te^t/t).</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 simplify first: d/dt (te^t/t) = d/dt (e^t)</p>
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<p>To differentiate the function, we simplify first: d/dt (te^t/t) = d/dt (e^t)</p>
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<p>The derivative of e^t is e^t. Therefore, d/dt (te^t/t) = e^t</p>
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<p>The derivative of e^t is e^t. Therefore, d/dt (te^t/t) = e^t</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we first simplify the given function to a more straightforward form and then differentiate. The simplification helps avoid unnecessary complexity in differentiation.</p>
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<p>In this process, we first simplify the given function to a more straightforward form and then differentiate. The simplification helps avoid unnecessary complexity in differentiation.</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 te^t</h2>
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<h2>FAQs on the Derivative of te^t</h2>
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<h3>1.Find the derivative of te^t.</h3>
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<h3>1.Find the derivative of te^t.</h3>
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<p>Using the product rule for te^t gives: d/dt (te^t) = e^t + te^t</p>
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<p>Using the product rule for te^t gives: d/dt (te^t) = e^t + te^t</p>
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<h3>2.Can we use the derivative of te^t in real life?</h3>
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<h3>2.Can we use the derivative of te^t in real life?</h3>
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<p>Yes, the derivative of te^t can model growth processes, such as population growth or financial growth, where the rate of change is dependent on time.</p>
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<p>Yes, the derivative of te^t can model growth processes, such as population growth or financial growth, where the rate of change is dependent on time.</p>
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<h3>3.Is the derivative of te^t defined for all t?</h3>
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<h3>3.Is the derivative of te^t defined for all t?</h3>
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<p>Yes, the derivative of te^t is defined for all real<a>numbers</a>t.</p>
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<p>Yes, the derivative of te^t is defined for all real<a>numbers</a>t.</p>
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<h3>4.What rule is used to differentiate te^t?</h3>
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<h3>4.What rule is used to differentiate te^t?</h3>
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<p>The product rule is used to differentiate te^t since it is a product of t and e^t.</p>
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<p>The product rule is used to differentiate te^t since it is a product of t and e^t.</p>
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<h3>5.Are the derivatives of te^t and e^t the same?</h3>
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<h3>5.Are the derivatives of te^t and e^t the same?</h3>
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<p>No, they are different. The derivative of te^t is e^t + te^t, while the derivative of e^t is just e^t.</p>
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<p>No, they are different. The derivative of te^t is e^t + te^t, while the derivative of e^t is just e^t.</p>
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<h2>Important Glossaries for the Derivative of te^t</h2>
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<h2>Important Glossaries for the Derivative of te^t</h2>
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<ul><li><strong>Derivative:</strong>Indicates how a function changes in response to a slight change in its input variable. </li>
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<ul><li><strong>Derivative:</strong>Indicates how a function changes in response to a slight change in its input variable. </li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form e^t, 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^t, where e is the base of natural logarithms. </li>
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</ul><ul><li><strong>Product Rule:</strong>A differentiation rule used to differentiate the product of two functions. </li>
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</ul><ul><li><strong>Product Rule:</strong>A differentiation rule used to differentiate the product of two functions. </li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives obtained by differentiating a function multiple times. </li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives obtained by differentiating a function multiple times. </li>
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</ul><ul><li><strong>First Principle:</strong>A fundamental method of finding derivatives using the limit of the difference quotient.</li>
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</ul><ul><li><strong>First Principle:</strong>A fundamental method of finding derivatives using the limit of the difference quotient.</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>