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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^-5x to understand how the function changes in response to a slight change in x. Derivatives are crucial in various real-life scenarios, such as calculating profit or loss. We will now discuss the derivative of e^-5x in detail.</p>
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<p>We use the derivative of e^-5x to understand how the function changes in response to a slight change in x. Derivatives are crucial in various real-life scenarios, such as calculating profit or loss. We will now discuss the derivative of e^-5x in detail.</p>
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<h2>What is the Derivative of e^-5x?</h2>
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<h2>What is the Derivative of e^-5x?</h2>
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<p>We now understand the derivative<a>of</a>e^-5x. It is commonly represented as d/dx (e^-5x) or (e^-5x)', and its value is -5e^-5x. </p>
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<p>We now understand the derivative<a>of</a>e^-5x. It is commonly represented as d/dx (e^-5x) or (e^-5x)', and its value is -5e^-5x. </p>
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<p>The<a>function</a>e^-5x has a clearly defined derivative, indicating it is differentiable everywhere. The key concepts are mentioned below:</p>
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<p>The<a>function</a>e^-5x has a clearly defined derivative, indicating it is differentiable everywhere. The key concepts are mentioned below:</p>
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<p>Exponential Function: e^-5x is an exponential function with a<a>base</a>e and a<a>constant</a><a>exponent</a>.</p>
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<p>Exponential Function: e^-5x is an exponential function with a<a>base</a>e and a<a>constant</a><a>exponent</a>.</p>
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<p>Chain Rule: A rule for differentiating compositions of functions, which is often used when differentiating exponential functions.</p>
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<p>Chain Rule: A rule for differentiating compositions of functions, which is often used when differentiating exponential functions.</p>
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<p>Constant Multiplier Rule: A rule used to differentiate functions multiplied by a constant.</p>
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<p>Constant Multiplier Rule: A rule used to differentiate functions multiplied by a constant.</p>
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<h2>Derivative of e^-5x Formula</h2>
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<h2>Derivative of e^-5x Formula</h2>
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<p>The derivative of e^-5x can be denoted as d/dx (e^-5x) or (e^-5x)'. The<a>formula</a>we use to differentiate e^-5x is: d/dx (e^-5x) = -5e^-5x The formula applies to all x.</p>
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<p>The derivative of e^-5x can be denoted as d/dx (e^-5x) or (e^-5x)'. The<a>formula</a>we use to differentiate e^-5x is: d/dx (e^-5x) = -5e^-5x The formula applies to all x.</p>
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<h2>Proofs of the Derivative of e^-5x</h2>
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<h2>Proofs of the Derivative of e^-5x</h2>
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<p>We can derive the derivative of e^-5x using several methods. </p>
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<p>We can derive the derivative of e^-5x using several methods. </p>
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<p>To show this, we will use the chain rule and the properties of exponential functions.</p>
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<p>To show this, we will use the chain rule and the properties of exponential functions.</p>
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<p>Here are the methods we use to prove this: Using Chain Rule To prove the differentiation of e^-5x using the chain rule, Consider u(x) = -5x and the outer function v(u) = e^u.</p>
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<p>Here are the methods we use to prove this: Using Chain Rule To prove the differentiation of e^-5x using the chain rule, Consider u(x) = -5x and the outer function v(u) = e^u.</p>
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<p>The derivative of v with respect to u is v'(u) = e^u, and the derivative of u with respect to x is u'(x) = -5. Using the chain rule: (v(u(x)))' = v'(u(x)) * u'(x), d/dx (e^-5x) = e^-5x * (-5) = -5e^-5x.</p>
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<p>The derivative of v with respect to u is v'(u) = e^u, and the derivative of u with respect to x is u'(x) = -5. Using the chain rule: (v(u(x)))' = v'(u(x)) * u'(x), d/dx (e^-5x) = e^-5x * (-5) = -5e^-5x.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using Constant Multiplier Rule We will now prove the derivative of e^-5x using the constant<a>multiplier</a>rule.</p>
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<p>Using Constant Multiplier Rule We will now prove the derivative of e^-5x using the constant<a>multiplier</a>rule.</p>
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<p>Given that d/dx (e^x) = e^x, For y = e^-5x, let u = -5x, then dy/du = e^u and du/dx = -5.</p>
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<p>Given that d/dx (e^x) = e^x, For y = e^-5x, let u = -5x, then dy/du = e^u and du/dx = -5.</p>
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<p>By the chain rule, dy/dx = dy/du * du/dx = e^u * (-5) = -5e^-5x. Thus, the derivative of e^-5x is -5e^-5x.</p>
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<p>By the chain rule, dy/dx = dy/du * du/dx = e^u * (-5) = -5e^-5x. Thus, the derivative of e^-5x is -5e^-5x.</p>
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<h2>Higher-Order Derivatives of e^-5x</h2>
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<h2>Higher-Order Derivatives of e^-5x</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.</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.</p>
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<p>Higher-order derivatives make it easier to understand functions like e^-5x. 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 make it easier to understand functions like e^-5x. 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, which is denoted using f′′(x).</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x).</p>
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<p>Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>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^-5x, the pattern can be observed as fⁿ(x) = (-5)ⁿe^-5x, indicating the change in the rate of change.</p>
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<p>For the nth derivative of e^-5x, the pattern can be observed as fⁿ(x) = (-5)ⁿe^-5x, indicating 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 x approaches infinity, e^-5x approaches zero, meaning the derivative -5e^-5x also approaches zero. When x is 0, the derivative of e^-5x = -5e^0, which is -5.</p>
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<p>When x approaches infinity, e^-5x approaches zero, meaning the derivative -5e^-5x also approaches zero. When x is 0, the derivative of e^-5x = -5e^0, which is -5.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^-5x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^-5x</h2>
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<p>Students frequently make mistakes when differentiating e^-5x. 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^-5x. 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^-5x · sin(x).</p>
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<p>Calculate the derivative of e^-5x · sin(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^-5x · sin(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-5x and v = sin(x). </p>
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<p>Here, we have f(x) = e^-5x · sin(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-5x and v = sin(x). </p>
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<p>Let’s differentiate each term, u′ = d/dx (e^-5x) = -5e^-5x v′ = d/dx (sin(x)) = cos(x) Substituting into the given equation, f'(x) = (-5e^-5x) · sin(x) + (e^-5x) · cos(x)</p>
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<p>Let’s differentiate each term, u′ = d/dx (e^-5x) = -5e^-5x v′ = d/dx (sin(x)) = cos(x) Substituting into the given equation, f'(x) = (-5e^-5x) · sin(x) + (e^-5x) · cos(x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -5e^-5x sin(x) + e^-5x cos(x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -5e^-5x sin(x) + e^-5x cos(x)</p>
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<p>Thus, the derivative of the specified function is -5e^-5x sin(x) + e^-5x cos(x).</p>
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<p>Thus, the derivative of the specified function is -5e^-5x sin(x) + e^-5x cos(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 uses the function y = e^-5x to model the decay of a chemical substance. Determine the rate of decay when x = 1.</p>
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<p>A company uses the function y = e^-5x to model the decay of a chemical substance. Determine 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^-5x (rate of decay)...(1) Now, we will differentiate the equation (1) Take the derivative of e^-5x: dy/dx = -5e^-5x</p>
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<p>We have y = e^-5x (rate of decay)...(1) Now, we will differentiate the equation (1) Take the derivative of e^-5x: dy/dx = -5e^-5x</p>
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<p>Given x = 1 (substitute this into the derivative) dy/dx = -5e^-5(1) = -5e^-5 Hence, the rate of decay of the chemical substance when x = 1 is -5e^-5.</p>
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<p>Given x = 1 (substitute this into the derivative) dy/dx = -5e^-5(1) = -5e^-5 Hence, the rate of decay of the chemical substance when x = 1 is -5e^-5.</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 substituting into the derivative. This indicates how quickly the substance is decaying at that specific point.</p>
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<p>We find the rate of decay at x = 1 by substituting into the derivative. This indicates how quickly the substance is decaying at that specific 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^-5x.</p>
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<p>Derive the second derivative of the function y = e^-5x.</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 = -5e^-5x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-5e^-5x] = -5 d/dx [e^-5x] = -5(-5e^-5x) = 25e^-5x Therefore, the second derivative of the function y = e^-5x is 25e^-5x.</p>
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<p>The first step is to find the first derivative, dy/dx = -5e^-5x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-5e^-5x] = -5 d/dx [e^-5x] = -5(-5e^-5x) = 25e^-5x Therefore, the second derivative of the function y = e^-5x is 25e^-5x.</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. </p>
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<p>We use the step-by-step process, where we start with the first derivative. </p>
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<p>We then differentiate the first derivative to find the second derivative.</p>
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<p>We then differentiate the first derivative 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^-5x)^2 = -10e^-10x.</p>
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<p>Prove: d/dx (e^-5x)^2 = -10e^-10x.</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 chain rule: Consider y = (e^-5x)^2 [e^-5x]^2 To differentiate, we use the chain rule: dy/dx = 2e^-5x · d/dx [e^-5x]</p>
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<p>Let’s start using the chain rule: Consider y = (e^-5x)^2 [e^-5x]^2 To differentiate, we use the chain rule: dy/dx = 2e^-5x · d/dx [e^-5x]</p>
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<p>Since the derivative of e^-5x is -5e^-5x, dy/dx = 2e^-5x · (-5e^-5x) = -10e^-10x Hence proved.</p>
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<p>Since the derivative of e^-5x is -5e^-5x, dy/dx = 2e^-5x · (-5e^-5x) = -10e^-10x 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 chain rule to differentiate the equation. </p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. </p>
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<p>Then, we replace e^-5x with its derivative. As a final step, we substitute back to derive the equation.</p>
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<p>Then, we replace e^-5x with its derivative. As a final step, we substitute back 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^-5x/x).</p>
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<p>Solve: d/dx (e^-5x/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^-5x/x) = (d/dx (e^-5x) · x - e^-5x · d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (e^-5x/x) = (d/dx (e^-5x) · x - e^-5x · d/dx(x))/x²</p>
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<p>We will substitute d/dx (e^-5x) = -5e^-5x and d/dx(x) = 1 (-5e^-5x · x - e^-5x · 1) / x² = (-5xe^-5x - e^-5x) / x² = -e^-5x(5x + 1) / x² Therefore, d/dx (e^-5x/x) = -e^-5x(5x + 1) / x²</p>
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<p>We will substitute d/dx (e^-5x) = -5e^-5x and d/dx(x) = 1 (-5e^-5x · x - e^-5x · 1) / x² = (-5xe^-5x - e^-5x) / x² = -e^-5x(5x + 1) / x² Therefore, d/dx (e^-5x/x) = -e^-5x(5x + 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. </p>
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<p>In this process, we differentiate the given function using the quotient rule. </p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>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^-5x</h2>
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<h2>FAQs on the Derivative of e^-5x</h2>
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<h3>1.Find the derivative of e^-5x.</h3>
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<h3>1.Find the derivative of e^-5x.</h3>
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<p>Using the chain rule for e^-5x, we differentiate as d/dx (e^-5x) = -5e^-5x.</p>
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<p>Using the chain rule for e^-5x, we differentiate as d/dx (e^-5x) = -5e^-5x.</p>
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<h3>2.Can we use the derivative of e^-5x in real life?</h3>
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<h3>2.Can we use the derivative of e^-5x in real life?</h3>
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<p>Yes, we can use the derivative of e^-5x in real life, especially in fields such as physics, chemistry, and finance, to model decay processes and calculate rates of change.</p>
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<p>Yes, we can use the derivative of e^-5x in real life, especially in fields such as physics, chemistry, and finance, to model decay processes and calculate rates of change.</p>
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<h3>3.Is it possible to take the derivative of e^-5x at any point?</h3>
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<h3>3.Is it possible to take the derivative of e^-5x at any point?</h3>
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<p>Yes, the function e^-5x is differentiable at all points, so it is always possible to take the derivative.</p>
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<p>Yes, the function e^-5x is differentiable at all points, so it is always possible to take the derivative.</p>
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<h3>4.What rule is used to differentiate e^-5x/x?</h3>
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<h3>4.What rule is used to differentiate e^-5x/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate e^-5x/x, d/dx (e^-5x/x) = (-5xe^-5x - e^-5x) / x².</p>
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<p>We use the<a>quotient</a>rule to differentiate e^-5x/x, d/dx (e^-5x/x) = (-5xe^-5x - e^-5x) / x².</p>
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<h3>5.Are the derivatives of e^-5x and e^5x the same?</h3>
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<h3>5.Are the derivatives of e^-5x and e^5x the same?</h3>
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<p>No, they are different. The derivative of e^-5x is -5e^-5x, while the derivative of e^5x is 5e^5x.</p>
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<p>No, they are different. The derivative of e^-5x is -5e^-5x, while the derivative of e^5x is 5e^5x.</p>
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<h2>Important Glossaries for the Derivative of e^-5x</h2>
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<h2>Important Glossaries for the Derivative of e^-5x</h2>
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<ul><li>Derivative: 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>Derivative: 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>Exponential Function: A mathematical function in which an independent variable appears in the exponent.</li>
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</ul><ul><li>Exponential Function: A mathematical function in which an independent variable appears in the exponent.</li>
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</ul><ul><li>Chain Rule: A rule for differentiating compositions of functions.</li>
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</ul><ul><li>Chain Rule: A rule for differentiating compositions of functions.</li>
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</ul><ul><li>Constant Multiplier Rule: A rule for differentiating functions multiplied by a constant.</li>
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</ul><ul><li>Constant Multiplier Rule: A rule for differentiating functions multiplied by a constant.</li>
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</ul><ul><li>Quotient Rule: A rule for differentiating a quotient of two functions.</li>
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</ul><ul><li>Quotient Rule: A rule for differentiating 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>