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2026-01-01
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>We use the derivative of e^-3x, which is -3e^-3x, to understand how the exponential function changes with a small change in x. Derivatives help us in various applications, including calculating rates of change in scientific and engineering contexts. We will now discuss the derivative of e^-3x in detail.</p>
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<p>We use the derivative of e^-3x, which is -3e^-3x, to understand how the exponential function changes with a small change in x. Derivatives help us in various applications, including calculating rates of change in scientific and engineering contexts. We will now discuss the derivative of e^-3x in detail.</p>
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<h2>What is the Derivative of e^-3x?</h2>
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<h2>What is the Derivative of e^-3x?</h2>
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<p>We now understand the derivative of e^-3x. It is commonly represented as d/dx (e^-3x) or (e^-3x)', and its value is -3e^-3x. The<a>function</a>e^-3x has a well-defined derivative, indicating it is differentiable for all<a>real numbers</a>.</p>
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<p>We now understand the derivative of e^-3x. It is commonly represented as d/dx (e^-3x) or (e^-3x)', and its value is -3e^-3x. The<a>function</a>e^-3x has a well-defined derivative, indicating it is differentiable for all<a>real numbers</a>.</p>
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<p>The key concepts are mentioned below:</p>
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<p>The key concepts are mentioned below:</p>
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<p><strong>Exponential Function:</strong>e^x, where e is the<a>base</a>of the natural logarithm.</p>
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<p><strong>Exponential Function:</strong>e^x, where e is the<a>base</a>of the natural logarithm.</p>
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<p><strong>Chain Rule:</strong>A rule used for differentiating compositions<a>of functions</a>.</p>
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<p><strong>Chain Rule:</strong>A rule used for differentiating compositions<a>of functions</a>.</p>
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<p><strong>Negative Exponent:</strong>Represents the reciprocal of the base raised to the positive<a>exponent</a>.</p>
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<p><strong>Negative Exponent:</strong>Represents the reciprocal of the base raised to the positive<a>exponent</a>.</p>
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<h2>Derivative of e^-3x Formula</h2>
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<h2>Derivative of e^-3x Formula</h2>
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<p>The derivative of e^-3x can be denoted as d/dx (e^-3x) or (e^-3x)'.</p>
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<p>The derivative of e^-3x can be denoted as d/dx (e^-3x) or (e^-3x)'.</p>
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<p>The<a>formula</a>we use to differentiate e^-3x is: d/dx (e^-3x) = -3e^-3x</p>
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<p>The<a>formula</a>we use to differentiate e^-3x is: d/dx (e^-3x) = -3e^-3x</p>
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<p>This formula holds for all x.</p>
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<p>This formula holds for all x.</p>
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<h2>Proofs of the Derivative of e^-3x</h2>
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<h2>Proofs of the Derivative of e^-3x</h2>
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<p>We can derive the derivative of e^-3x using different proofs. To demonstrate this, we will use differentiation rules, particularly focusing on the chain rule.</p>
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<p>We can derive the derivative of e^-3x using different proofs. To demonstrate this, we will use differentiation rules, particularly focusing on the chain rule.</p>
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<p>Here are some methods to prove this derivative: By Chain Rule To prove the differentiation of e^-3x</p>
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<p>Here are some methods to prove this derivative: By Chain Rule To prove the differentiation of e^-3x</p>
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<p>using the chain rule: Consider f(x) = e^u where u = -3x.</p>
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<p>using the chain rule: Consider f(x) = e^u where u = -3x.</p>
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<p>The derivative of e^u is e^u du/dx. Here, du/dx = -3. So, d/dx (e^-3x) = e^-3x * (-3) = -3e^-3x.</p>
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<p>The derivative of e^u is e^u du/dx. Here, du/dx = -3. So, d/dx (e^-3x) = e^-3x * (-3) = -3e^-3x.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h2>Higher-Order Derivatives of e^-3x</h2>
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<h2>Higher-Order Derivatives of e^-3x</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 can be complex. To understand them better, think of velocity (first derivative) and acceleration (second derivative) in physics.</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 can be complex. To understand them better, think of velocity (first derivative) and acceleration (second derivative) in physics.</p>
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<p>Higher-order derivatives help us analyze functions like e^-3x. For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.</p>
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<p>Higher-order derivatives help us analyze functions like e^-3x. For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.</p>
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<p>The second derivative is derived from the first derivative, 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>The second derivative is derived from the first derivative, 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^-3x, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the<a>rate</a>of change.</p>
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<p>For the nth derivative of e^-3x, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the<a>rate</a>of change.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>For all x, the derivative of e^-3x is defined. When x is a large<a>negative number</a>, the exponential function e^-3x approaches zero, and its derivative also approaches zero.</p>
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<p>For all x, the derivative of e^-3x is defined. When x is a large<a>negative number</a>, the exponential function e^-3x approaches zero, and its derivative also approaches zero.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^-3x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^-3x</h2>
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<p>Students frequently make mistakes when differentiating e^-3x. These mistakes can be avoided by understanding the correct methods. Here are some common mistakes and ways to solve them:</p>
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<p>Students frequently make mistakes when differentiating e^-3x. These mistakes can be avoided by understanding the correct methods. Here are some 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^-3x · x²)</p>
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<p>Calculate the derivative of (e^-3x · 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>Here, we have f(x) = e^-3x · x².</p>
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<p>Here, we have f(x) = e^-3x · x².</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-3x and v = x².</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-3x and v = x².</p>
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<p>Let's differentiate each term, u′ = d/dx (e^-3x) = -3e^-3x v′ = d/dx (x²) = 2x</p>
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<p>Let's differentiate each term, u′ = d/dx (e^-3x) = -3e^-3x v′ = d/dx (x²) = 2x</p>
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<p>Substituting into the given equation, f'(x) = (-3e^-3x) · x² + e^-3x · (2x)</p>
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<p>Substituting into the given equation, f'(x) = (-3e^-3x) · x² + e^-3x · (2x)</p>
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<p>Let's simplify terms to get the final answer, f'(x) = -3x²e^-3x + 2xe^-3x</p>
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<p>Let's simplify terms to get the final answer, f'(x) = -3x²e^-3x + 2xe^-3x</p>
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<p>Thus, the derivative of the specified function is -3x²e^-3x + 2xe^-3x.</p>
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<p>Thus, the derivative of the specified function is -3x²e^-3x + 2xe^-3x.</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 scientist is measuring the decay of a radioactive substance, represented by the function y = e^-3x, where y is the remaining quantity at time x. Calculate the rate of decay when x = 1.</p>
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<p>A scientist is measuring the decay of a radioactive substance, represented by the function y = e^-3x, where y is the remaining quantity at time x. Calculate the rate of decay 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 y = e^-3x (decay function)...(1)</p>
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<p>We have y = e^-3x (decay function)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of e^-3x: dy/dx = -3e^-3x</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of e^-3x: dy/dx = -3e^-3x</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>dy/dx = -3e^-3(1) = -3e^-3 The rate of decay at x = 1 is -3e^-3.</p>
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<p>dy/dx = -3e^-3(1) = -3e^-3 The rate of decay at x = 1 is -3e^-3.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of decay at x = 1 by taking the derivative of the decay function.</p>
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<p>We find the rate of decay at x = 1 by taking the derivative of the decay function.</p>
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<p>Substituting x = 1 into the derivative gives us the rate at that point.</p>
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<p>Substituting x = 1 into the derivative gives us the rate at that point.</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^-3x.</p>
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<p>Derive the second derivative of the function y = e^-3x.</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 = -3e^-3x...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = -3e^-3x...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3e^-3x]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3e^-3x]</p>
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<p>d²y/dx² = (-3) * (-3e^-3x)</p>
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<p>d²y/dx² = (-3) * (-3e^-3x)</p>
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<p>d²y/dx² = 9e^-3x</p>
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<p>d²y/dx² = 9e^-3x</p>
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<p>Therefore, the second derivative of the function y = e^-3x is 9e^-3x.</p>
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<p>Therefore, the second derivative of the function y = e^-3x is 9e^-3x.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use a step-by-step process, where we start with the first derivative. We then differentiate it again and simplify to find the second derivative.</p>
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<p>We use a step-by-step process, where we start with the first derivative. We then differentiate it again and simplify 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^-3x · e^x) = -2e^-2x.</p>
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<p>Prove: d/dx (e^-3x · e^x) = -2e^-2x.</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^-3x · e^x</p>
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<p>Let's start using the product rule: Consider y = e^-3x · e^x</p>
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<p>To differentiate, we use the product rule: dy/dx = (d/dx [e^-3x]) · e^x + e^-3x · (d/dx [e^x])</p>
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<p>To differentiate, we use the product rule: dy/dx = (d/dx [e^-3x]) · e^x + e^-3x · (d/dx [e^x])</p>
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<p>The derivatives are: d/dx [e^-3x] = -3e^-3x d/dx [e^x] = e^x</p>
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<p>The derivatives are: d/dx [e^-3x] = -3e^-3x d/dx [e^x] = e^x</p>
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<p>Substitute the derivatives: dy/dx = (-3e^-3x) · e^x + e^-3x · e^x = -3e^-2x + e^-2x = -2e^-2x</p>
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<p>Substitute the derivatives: dy/dx = (-3e^-3x) · e^x + e^-3x · e^x = -3e^-2x + e^-2x = -2e^-2x</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 then simplify by combining like terms 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 then simplify by combining like terms 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^-3x/x)</p>
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<p>Solve: d/dx (e^-3x/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^-3x/x) = (d/dx [e^-3x] · x - e^-3x · d/dx [x]) / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (e^-3x/x) = (d/dx [e^-3x] · x - e^-3x · d/dx [x]) / x²</p>
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<p>We will substitute d/dx [e^-3x] = -3e^-3x and d/dx [x] = 1 = (-3e^-3x · x - e^-3x · 1) / x² = (-3xe^-3x - e^-3x) / x² = -e^-3x(3x + 1) / x²</p>
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<p>We will substitute d/dx [e^-3x] = -3e^-3x and d/dx [x] = 1 = (-3e^-3x · x - e^-3x · 1) / x² = (-3xe^-3x - e^-3x) / x² = -e^-3x(3x + 1) / x²</p>
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<p>Therefore, d/dx (e^-3x/x) = -e^-3x(3x + 1) / x²</p>
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<p>Therefore, d/dx (e^-3x/x) = -e^-3x(3x + 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^-3x</h2>
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<h2>FAQs on the Derivative of e^-3x</h2>
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<h3>1.Find the derivative of e^-3x.</h3>
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<h3>1.Find the derivative of e^-3x.</h3>
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<p>Using the chain rule, where u = -3x, d/dx (e^-3x) = -3e^-3x.</p>
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<p>Using the chain rule, where u = -3x, d/dx (e^-3x) = -3e^-3x.</p>
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<h3>2.Can we use the derivative of e^-3x in real life?</h3>
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<h3>2.Can we use the derivative of e^-3x in real life?</h3>
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<p>Yes, we can use the derivative of e^-3x in real life to model<a>exponential decay</a>processes, which are common in fields like physics, chemistry, and biology.</p>
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<p>Yes, we can use the derivative of e^-3x in real life to model<a>exponential decay</a>processes, which are common in fields like physics, chemistry, and biology.</p>
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<h3>3.What rule is used to differentiate e^-3x/x?</h3>
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<h3>3.What rule is used to differentiate e^-3x/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate e^-3x/x: d/dx (e^-3x/x) = (x · (-3e^-3x) - e^-3x · 1) / x².</p>
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<p>We use the<a>quotient</a>rule to differentiate e^-3x/x: d/dx (e^-3x/x) = (x · (-3e^-3x) - e^-3x · 1) / x².</p>
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<h3>4.Are the derivatives of e^-3x and e^3x the same?</h3>
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<h3>4.Are the derivatives of e^-3x and e^3x the same?</h3>
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<p>No, they are different. The derivative of e^-3x is -3e^-3x, while the derivative of e^3x is 3e^3x.</p>
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<p>No, they are different. The derivative of e^-3x is -3e^-3x, while the derivative of e^3x is 3e^3x.</p>
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<h3>5.What is the second derivative of e^-3x?</h3>
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<h3>5.What is the second derivative of e^-3x?</h3>
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<p>The second derivative of e^-3x is 9e^-3x, obtained by differentiating the first derivative -3e^-3x.</p>
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<p>The second derivative of e^-3x is 9e^-3x, obtained by differentiating the first derivative -3e^-3x.</p>
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<h2>Important Glossaries for the Derivative of e^-3x</h2>
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<h2>Important Glossaries for the Derivative of e^-3x</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 small 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 small change in x.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A mathematical function in the form of e^x, where e is the base of natural logarithms.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A mathematical function in the form of e^x, where e is the base of natural logarithms.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating the composition of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating the composition of 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>Quotient Rule:</strong>A rule used to differentiate the quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate 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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<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>