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2026-01-01
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<p>Last updated on<strong>September 1, 2025</strong></p>
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<p>Last updated on<strong>September 1, 2025</strong></p>
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<p>We use the derivative of e^y, which is e^y(dy/dx) when y is a function of x, as a tool to understand how the function changes with respect to a variable. Derivatives are crucial in various real-life applications including scientific modeling and economics. We will now discuss the derivative of e^y in detail.</p>
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<p>We use the derivative of e^y, which is e^y(dy/dx) when y is a function of x, as a tool to understand how the function changes with respect to a variable. Derivatives are crucial in various real-life applications including scientific modeling and economics. We will now discuss the derivative of e^y in detail.</p>
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<h2>What is the Derivative of e^y?</h2>
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<h2>What is the Derivative of e^y?</h2>
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<p>The derivative<a>of</a>ey is an important concept in<a>calculus</a>.</p>
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<p>The derivative<a>of</a>ey is an important concept in<a>calculus</a>.</p>
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<p>It is generally represented as d/dx(ey) or (ey)'.</p>
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<p>It is generally represented as d/dx(ey) or (ey)'.</p>
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<p>Since e^y is an exponential<a>function</a>, its derivative is itself multiplied by the derivative of its<a>exponent</a>. The key concepts involved are:</p>
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<p>Since e^y is an exponential<a>function</a>, its derivative is itself multiplied by the derivative of its<a>exponent</a>. The key concepts involved are:</p>
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<p>Exponential Function: ey where e is a<a>constant</a>approximately equal to 2.71828.</p>
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<p>Exponential Function: ey where e is a<a>constant</a>approximately equal to 2.71828.</p>
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<p>Chain Rule: Used for differentiating composite functions like ey when y is a function of x. Implicit</p>
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<p>Chain Rule: Used for differentiating composite functions like ey when y is a function of x. Implicit</p>
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<p>Differentiation: Necessary when y is implicitly defined in<a>terms</a>of x.</p>
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<p>Differentiation: Necessary when y is implicitly defined in<a>terms</a>of x.</p>
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<h2>Derivative of e^y Formula</h2>
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<h2>Derivative of e^y Formula</h2>
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<p>The derivative of ey, when y is a function of x, can be denoted as d/dx(ey) or (ey)'.</p>
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<p>The derivative of ey, when y is a function of x, can be denoted as d/dx(ey) or (ey)'.</p>
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<p>The<a>formula</a>used to differentiate ey is: d/dx(ey) = ey(dy/dx)</p>
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<p>The<a>formula</a>used to differentiate ey is: d/dx(ey) = ey(dy/dx)</p>
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<p>This formula is applicable whenever y is a differentiable function of x.</p>
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<p>This formula is applicable whenever y is a differentiable function of x.</p>
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<h2>Proofs of the Derivative of e^y</h2>
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<h2>Proofs of the Derivative of e^y</h2>
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<p>We can derive the derivative of e^y using different methods.</p>
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<p>We can derive the derivative of e^y using different methods.</p>
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<p>To show this, we will apply the chain rule and the concept of implicit differentiation.</p>
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<p>To show this, we will apply the chain rule and the concept of implicit differentiation.</p>
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<p>Here are the methods used to prove this:</p>
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<p>Here are the methods used to prove this:</p>
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<p>Using Chain Rule To prove the differentiation of ey using the chain rule, consider y as a function of x:</p>
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<p>Using Chain Rule To prove the differentiation of ey using the chain rule, consider y as a function of x:</p>
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<p>Let u = y, where y is a function of x.</p>
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<p>Let u = y, where y is a function of x.</p>
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<p>Then, ey = eu.</p>
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<p>Then, ey = eu.</p>
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<p>Applying the chain rule, we get: d/dx(ey) = d/dx(eu) = eu * du/dx Since u = y, du/dx = dy/dx.</p>
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<p>Applying the chain rule, we get: d/dx(ey) = d/dx(eu) = eu * du/dx Since u = y, du/dx = dy/dx.</p>
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<p>Therefore, d/dx(ey) = ey * dy/dx.</p>
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<p>Therefore, d/dx(ey) = ey * dy/dx.</p>
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<p>Using Implicit Differentiation Assume y is implicitly defined in terms of x.</p>
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<p>Using Implicit Differentiation Assume y is implicitly defined in terms of x.</p>
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<p>Differentiate both sides of the<a>equation</a>ey = z with respect to x: d/dx(ey) = d/dx(z)</p>
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<p>Differentiate both sides of the<a>equation</a>ey = z with respect to x: d/dx(ey) = d/dx(z)</p>
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<p>Using the chain rule, we get: ey * dy/dx = dz/dx</p>
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<p>Using the chain rule, we get: ey * dy/dx = dz/dx</p>
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<p>If z = e^y, then dz/dx = ey * dy/dx.</p>
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<p>If z = e^y, then dz/dx = ey * dy/dx.</p>
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<h2>Higher-Order Derivatives of e^y</h2>
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<h2>Higher-Order Derivatives of e^y</h2>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are known as higher-order derivatives.</p>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are known as higher-order derivatives.</p>
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<p>Higher-order derivatives can provide more insight into the behavior of the function.</p>
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<p>Higher-order derivatives can provide more insight into the behavior of the function.</p>
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<p>Consider the case of ey:</p>
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<p>Consider the case of ey:</p>
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<p>First Derivative: d/dx(ey) = ey(dy/dx)</p>
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<p>First Derivative: d/dx(ey) = ey(dy/dx)</p>
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<p>Second Derivative: Differentiate the first derivative again, using the<a>product</a>rule: d²/dx²(ey) = d/dx(ey(dy/dx)) = (ey(dy/dx))(dy/dx) + ey(d²y/dx²)</p>
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<p>Second Derivative: Differentiate the first derivative again, using the<a>product</a>rule: d²/dx²(ey) = d/dx(ey(dy/dx)) = (ey(dy/dx))(dy/dx) + ey(d²y/dx²)</p>
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<p>This process can be continued for higher-order derivatives.</p>
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<p>This process can be continued 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 y is constant, the derivative ey is simply ey multiplied by 0, which is 0. When y is a linear function of x, say y = ax, the derivative simplifies to ey * a.</p>
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<p>When y is constant, the derivative ey is simply ey multiplied by 0, which is 0. When y is a linear function of x, say y = ax, the derivative simplifies to ey * a.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^y</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of e^y</h2>
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<p>Students often make errors when differentiating ey, especially when y is a function of x. These mistakes can be corrected by understanding the correct process. Here are some common mistakes and how to solve them:</p>
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<p>Students often make errors when differentiating ey, especially when y is a function of x. These mistakes can be corrected by understanding the correct process. Here are some common mistakes and how 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^(2y) with respect to x, given y = x² + 1.</p>
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<p>Calculate the derivative of e^(2y) with respect to x, given y = x² + 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>Here, we have f(y) = e(2y) and y = x² + 1.</p>
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<p>Here, we have f(y) = e(2y) and y = x² + 1.</p>
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<p>Using the chain rule: d/dx(e(2y)) = e(2y) * d/dx(2y)</p>
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<p>Using the chain rule: d/dx(e(2y)) = e(2y) * d/dx(2y)</p>
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<p>Since y = x² + 1, we have: d/dx(2y) = 2 * d/dx(y) = 2 * 2x = 4x</p>
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<p>Since y = x² + 1, we have: d/dx(2y) = 2 * d/dx(y) = 2 * 2x = 4x</p>
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<p>So, the derivative is: f'(x) = e(2y) * 4x = 4x * e(2(x²+1))</p>
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<p>So, the derivative is: f'(x) = e(2y) * 4x = 4x * e(2(x²+1))</p>
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<p>Thus, the derivative of e(2y) with respect to x is 4x * e(2(x²+1)).</p>
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<p>Thus, the derivative of e(2y) with respect to x is 4x * e(2(x²+1)).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We differentiate e(2y) using the chain rule. By substituting y = x² + 1 and differentiating, we obtain the derivative.</p>
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<p>We differentiate e(2y) using the chain rule. By substituting y = x² + 1 and differentiating, we obtain the 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 observes growth modeled by e^y, where y is the profit function y = 3x + 2. Find the rate of change of growth at x = 1.</p>
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<p>A company observes growth modeled by e^y, where y is the profit function y = 3x + 2. Find the rate of change of growth at 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 ey as the growth model and y = 3x + 2.</p>
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<p>We have ey as the growth model and y = 3x + 2.</p>
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<p>First, differentiate y: dy/dx = 3</p>
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<p>First, differentiate y: dy/dx = 3</p>
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<p>Now, differentiate e^y: d/dx(ey) = ey * dy/dx = ey * 3</p>
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<p>Now, differentiate e^y: d/dx(ey) = ey * dy/dx = ey * 3</p>
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<p>At x = 1, y = 3(1) + 2 = 5</p>
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<p>At x = 1, y = 3(1) + 2 = 5</p>
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<p>Therefore, the rate of change of growth is: 3 * e5</p>
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<p>Therefore, the rate of change of growth is: 3 * e5</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The company's growth is modeled by ey, where y is profit. The rate of change is found by differentiating and substituting x = 1 into the equation.</p>
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<p>The company's growth is modeled by ey, where y is profit. The rate of change is found by differentiating and substituting x = 1 into 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 3</h3>
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<h3>Problem 3</h3>
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<p>Find the second derivative of e^y where y = ln(x).</p>
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<p>Find the second derivative of e^y where y = ln(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>First, find the first derivative: d/dx(ey) = ey * dy/dx</p>
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<p>First, find the first derivative: d/dx(ey) = ey * dy/dx</p>
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<p>Since y = ln(x), dy/dx = 1/x.</p>
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<p>Since y = ln(x), dy/dx = 1/x.</p>
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<p>Therefore, d/dx(e^y) = ey * 1/x</p>
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<p>Therefore, d/dx(e^y) = ey * 1/x</p>
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<p>Now, to find the second derivative: d²/dx²(ey) = d/dx(ey * 1/x)</p>
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<p>Now, to find the second derivative: d²/dx²(ey) = d/dx(ey * 1/x)</p>
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<p>Using the product rule: = (1/x) * d/dx(ey) + ey * d/dx(1/x) = (1/x) * (ey * 1/x) + ey * (-1/x²) = ey/x² - ey/x² = 0</p>
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<p>Using the product rule: = (1/x) * d/dx(ey) + ey * d/dx(1/x) = (1/x) * (ey * 1/x) + ey * (-1/x²) = ey/x² - ey/x² = 0</p>
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<p>So, the second derivative is 0.</p>
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<p>So, the second derivative is 0.</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 using implicit differentiation and apply the product rule for the second derivative. The terms simplify to zero.</p>
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<p>We start with the first derivative using implicit differentiation and apply the product rule for the second derivative. The terms simplify to zero.</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>Show that d/dx(e^(3y)) = 3e^(3y) * dy/dx.</p>
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<p>Show that d/dx(e^(3y)) = 3e^(3y) * dy/dx.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Using the chain rule:</p>
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<p>Using the chain rule:</p>
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<p>Consider f(x) = e(3y) d/dx(e(3y)) = e(3y) * d/dx(3y)</p>
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<p>Consider f(x) = e(3y) d/dx(e(3y)) = e(3y) * d/dx(3y)</p>
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<p>Since d/dx(3y) = 3 * dy/dx,</p>
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<p>Since d/dx(3y) = 3 * dy/dx,</p>
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<p>We have: d/dx(e(3y)) = 3e(3y) * dy/dx</p>
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<p>We have: d/dx(e(3y)) = 3e(3y) * dy/dx</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>We use the chain rule to differentiate e(3y). The derivative of the exponent is multiplied by the derivative of y, confirming the result.</p>
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<p>We use the chain rule to differentiate e(3y). The derivative of the exponent is multiplied by the derivative of y, confirming the result.</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>Differentiate e^(y/x) with respect to x.</p>
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<p>Differentiate e^(y/x) with respect to 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>Use the chain rule and quotient rule:</p>
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<p>Use the chain rule and quotient rule:</p>
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<p>d/dx(e(y/x)) = e(y/x) * d/dx(y/x)</p>
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<p>d/dx(e(y/x)) = e(y/x) * d/dx(y/x)</p>
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<p>Apply the quotient rule: d/dx(y/x) = (x * dy/dx - y * 1)/x² = (x * dy/dx - y)/x²</p>
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<p>Apply the quotient rule: d/dx(y/x) = (x * dy/dx - y * 1)/x² = (x * dy/dx - y)/x²</p>
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<p>Therefore, the derivative is: e(y/x) * ((x * dy/dx - y)/x²)</p>
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<p>Therefore, the derivative is: e(y/x) * ((x * dy/dx - y)/x²)</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We apply the chain rule and quotient rule to find the derivative of e(y/x) by differentiating the exponential function and its exponent.</p>
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<p>We apply the chain rule and quotient rule to find the derivative of e(y/x) by differentiating the exponential function and its exponent.</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^y</h2>
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<h2>FAQs on the Derivative of e^y</h2>
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<h3>1.What is the derivative of e^y?</h3>
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<h3>1.What is the derivative of e^y?</h3>
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<p>The derivative of ey with respect to x is ey * dy/dx, assuming y is a function of x.</p>
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<p>The derivative of ey with respect to x is ey * dy/dx, assuming y is a function of x.</p>
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<h3>2.How is the derivative of e^y used in real life?</h3>
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<h3>2.How is the derivative of e^y used in real life?</h3>
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<p>The derivative of ey is used in various fields such as physics for modeling<a>exponential growth</a>or decay, in economics for growth rates, and in engineering for signal processing.</p>
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<p>The derivative of ey is used in various fields such as physics for modeling<a>exponential growth</a>or decay, in economics for growth rates, and in engineering for signal processing.</p>
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<h3>3.Can we differentiate e^y when y is a constant?</h3>
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<h3>3.Can we differentiate e^y when y is a constant?</h3>
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<p>Yes, if y is a constant, d/dx(ey) = 0 because the derivative of a constant is zero.</p>
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<p>Yes, if y is a constant, d/dx(ey) = 0 because the derivative of a constant is zero.</p>
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<h3>4.What rule is used to differentiate e^(y/x)?</h3>
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<h3>4.What rule is used to differentiate e^(y/x)?</h3>
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<p>We use both the chain rule and the<a>quotient</a>rule to differentiate e(y/x).</p>
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<p>We use both the chain rule and the<a>quotient</a>rule to differentiate e(y/x).</p>
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<h3>5.Are the derivatives of e^y and e^x the same?</h3>
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<h3>5.Are the derivatives of e^y and e^x the same?</h3>
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<p>No, they are different. The derivative of ey is ey * dy/dx when y is a function of x, while the derivative of ex is simply ex.</p>
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<p>No, they are different. The derivative of ey is ey * dy/dx when y is a function of x, while the derivative of ex is simply ex.</p>
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<h2>Important Glossaries for the Derivative of e^y</h2>
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<h2>Important Glossaries for the Derivative of e^y</h2>
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<ul><li><strong>Derivative:</strong>A measure of how a function changes as its input changes, often represented as the slope of the tangent line to the curve of the function.</li>
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<ul><li><strong>Derivative:</strong>A measure of how a function changes as its input changes, often represented as the slope of the tangent line to the curve of the function.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form ey where e is the base of the natural logarithm, approximately equal to 2.71828.</li>
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</ul><ul><li><strong>Exponential Function:</strong>A function of the form ey where e is the base of the natural logarithm, approximately equal to 2.71828.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A method to find derivatives of functions that are not explicitly solved for one variable in terms of another.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A method to find derivatives of functions that are not explicitly solved for one variable in terms of another.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A technique for differentiating functions that are the ratio of two differentiable functions. </li>
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</ul><ul><li><strong>Quotient Rule:</strong>A technique for differentiating functions that are the ratio of two differentiable 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>