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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 f(x)^g(x) to understand how this function changes in response to a slight change in x. Derivatives help us calculate rates of change in real-life situations. We will now discuss the derivative of f(x)^g(x) in detail.</p>
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<p>We use the derivative of f(x)^g(x) to understand how this function changes in response to a slight change in x. Derivatives help us calculate rates of change in real-life situations. We will now discuss the derivative of f(x)^g(x) in detail.</p>
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<h2>What is the Derivative of f(x)^g(x)?</h2>
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<h2>What is the Derivative of f(x)^g(x)?</h2>
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<p>We now understand the derivative<a>of</a>f(x)^g(x). It is commonly represented as d/dx (f(x)^g(x)) and can be derived using logarithmic differentiation. The<a>function</a>f(x)^g(x) can be differentiable within its domain based on the functions f(x) and g(x). The key concepts are mentioned below: Logarithmic Differentiation: A method used to differentiate functions of the form f(x)^g(x). Chain Rule: Used in the differentiation process to handle composite functions. Product Rule: Applied when necessary to differentiate products of functions.</p>
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<p>We now understand the derivative<a>of</a>f(x)^g(x). It is commonly represented as d/dx (f(x)^g(x)) and can be derived using logarithmic differentiation. The<a>function</a>f(x)^g(x) can be differentiable within its domain based on the functions f(x) and g(x). The key concepts are mentioned below: Logarithmic Differentiation: A method used to differentiate functions of the form f(x)^g(x). Chain Rule: Used in the differentiation process to handle composite functions. Product Rule: Applied when necessary to differentiate products of functions.</p>
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<h2>Derivative of f(x)^g(x) Formula</h2>
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<h2>Derivative of f(x)^g(x) Formula</h2>
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<p>The derivative of f(x)^g(x) can be obtained using logarithmic differentiation. The<a>formula</a>is: d/dx (f(x)^g(x)) = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) This formula applies to all x where f(x) > 0 and both f(x) and g(x) are differentiable at x.</p>
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<p>The derivative of f(x)^g(x) can be obtained using logarithmic differentiation. The<a>formula</a>is: d/dx (f(x)^g(x)) = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) This formula applies to all x where f(x) > 0 and both f(x) and g(x) are differentiable at x.</p>
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<h2>Proofs of the Derivative of f(x)^g(x)</h2>
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<h2>Proofs of the Derivative of f(x)^g(x)</h2>
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<p>We can derive the derivative of f(x)^g(x) using proofs. To show this, we will use logarithmic differentiation along with the rules of differentiation. There are several methods we use to prove this, such as: By Logarithmic Differentiation Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of f(x)^g(x) results in the formula mentioned above using these methods: By Logarithmic Differentiation The derivative of f(x)^g(x) can be proved using logarithmic differentiation, which involves taking the natural logarithm of both sides of the function. To find the derivative of f(x)^g(x), we will consider y = f(x)^g(x). Taking the natural logarithm of both sides, we have: ln(y) = g(x)ln(f(x)) Differentiating both sides with respect to x, 1/y * dy/dx = g'(x)ln(f(x)) + g(x)f'(x)/f(x) Solving for dy/dx gives us: dy/dx = y * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Substituting y = f(x)^g(x), dy/dx = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Hence, proved. Using Chain Rule To prove the differentiation of f(x)^g(x) using the chain rule, We consider the function y = e^(g(x)ln(f(x))) Differentiating y using the chain rule: dy/dx = e^(g(x)ln(f(x))) * d/dx [g(x)ln(f(x))] The derivative of the<a>exponent</a>g(x)ln(f(x)) is: g'(x)ln(f(x)) + g(x)f'(x)/f(x) Therefore, dy/dx = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Using Product Rule We will now prove the derivative of f(x)^g(x) using the<a>product</a>rule. The step-by-step process is demonstrated below: Consider the function y = e^(g(x)ln(f(x))) Using the product rule, dy/dx = e^(g(x)ln(f(x))) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Simplifying, we get: dy/dx = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x))</p>
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<p>We can derive the derivative of f(x)^g(x) using proofs. To show this, we will use logarithmic differentiation along with the rules of differentiation. There are several methods we use to prove this, such as: By Logarithmic Differentiation Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of f(x)^g(x) results in the formula mentioned above using these methods: By Logarithmic Differentiation The derivative of f(x)^g(x) can be proved using logarithmic differentiation, which involves taking the natural logarithm of both sides of the function. To find the derivative of f(x)^g(x), we will consider y = f(x)^g(x). Taking the natural logarithm of both sides, we have: ln(y) = g(x)ln(f(x)) Differentiating both sides with respect to x, 1/y * dy/dx = g'(x)ln(f(x)) + g(x)f'(x)/f(x) Solving for dy/dx gives us: dy/dx = y * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Substituting y = f(x)^g(x), dy/dx = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Hence, proved. Using Chain Rule To prove the differentiation of f(x)^g(x) using the chain rule, We consider the function y = e^(g(x)ln(f(x))) Differentiating y using the chain rule: dy/dx = e^(g(x)ln(f(x))) * d/dx [g(x)ln(f(x))] The derivative of the<a>exponent</a>g(x)ln(f(x)) is: g'(x)ln(f(x)) + g(x)f'(x)/f(x) Therefore, dy/dx = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Using Product Rule We will now prove the derivative of f(x)^g(x) using the<a>product</a>rule. The step-by-step process is demonstrated below: Consider the function y = e^(g(x)ln(f(x))) Using the product rule, dy/dx = e^(g(x)ln(f(x))) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x)) Simplifying, we get: dy/dx = f(x)^g(x) * (g'(x)ln(f(x)) + g(x)f'(x)/f(x))</p>
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<h2>Higher-Order Derivatives of f(x)^g(x)</h2>
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<h2>Higher-Order Derivatives of f(x)^g(x)</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. To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like f(x)^g(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative and this pattern continues. For the nth Derivative of f(x)^g(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change continuing for higher-order derivatives.</p>
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<p>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. To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like f(x)^g(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative and this pattern continues. For the nth Derivative of f(x)^g(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change continuing for higher-order derivatives.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When f(x) = x^0, the derivative is zero because any<a>number</a>raised to the<a>power</a>of zero is 1, and the derivative of a<a>constant</a>is zero. When g(x) = 0, the derivative of f(x)^0 = 1, which is also zero, as it is a constant function.</p>
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<p>When f(x) = x^0, the derivative is zero because any<a>number</a>raised to the<a>power</a>of zero is 1, and the derivative of a<a>constant</a>is zero. When g(x) = 0, the derivative of f(x)^0 = 1, which is also zero, as it is a constant function.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of f(x)^g(x)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of f(x)^g(x)</h2>
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<p>Students frequently make mistakes when differentiating f(x)^g(x). 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 f(x)^g(x). 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 ((x^2 + 1)^x).</p>
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<p>Calculate the derivative of ((x^2 + 1)^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) = (x^2 + 1)^x. Using logarithmic differentiation, let y = (x^2 + 1)^x. Taking the natural logarithm of both sides: ln(y) = xln(x^2 + 1) Differentiating both sides, we get: 1/y * dy/dx = ln(x^2 + 1) + x * (2x/(x^2 + 1)) dy/dx = y * (ln(x^2 + 1) + 2x^2/(x^2 + 1)) Substituting y = (x^2 + 1)^x: dy/dx = (x^2 + 1)^x * (ln(x^2 + 1) + 2x^2/(x^2 + 1)) Thus, the derivative of the specified function is (x^2 + 1)^x * (ln(x^2 + 1) + 2x^2/(x^2 + 1)).</p>
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<p>Here, we have f(x) = (x^2 + 1)^x. Using logarithmic differentiation, let y = (x^2 + 1)^x. Taking the natural logarithm of both sides: ln(y) = xln(x^2 + 1) Differentiating both sides, we get: 1/y * dy/dx = ln(x^2 + 1) + x * (2x/(x^2 + 1)) dy/dx = y * (ln(x^2 + 1) + 2x^2/(x^2 + 1)) Substituting y = (x^2 + 1)^x: dy/dx = (x^2 + 1)^x * (ln(x^2 + 1) + 2x^2/(x^2 + 1)) Thus, the derivative of the specified function is (x^2 + 1)^x * (ln(x^2 + 1) + 2x^2/(x^2 + 1)).</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 logarithmic differentiation. The first step is taking the natural logarithm, differentiating it, and then using the product rule to get the final result.</p>
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<p>We find the derivative of the given function by applying logarithmic differentiation. The first step is taking the natural logarithm, differentiating it, and then 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 balloon is inflated such that its volume is represented by V = (3 + t)^t, where t is time in seconds. Find the rate at which the volume of the balloon is changing at t = 1 second.</p>
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<p>A balloon is inflated such that its volume is represented by V = (3 + t)^t, where t is time in seconds. Find the rate at which the volume of the balloon is changing at t = 1 second.</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 V = (3 + t)^t (volume of the balloon)...(1) Now, we will differentiate the equation (1) Take the derivative of (3 + t)^t: dV/dt = (3 + t)^t * (ln(3 + t) + t/(3 + t)) Given t = 1, substitute this into the derivative: dV/dt = (3 + 1)^1 * (ln(3 + 1) + 1/(3 + 1)) dV/dt = 4 * (ln(4) + 1/4) Calculate the value: dV/dt = 4 * (ln(4) + 0.25) Hence, we get the rate at which the volume of the balloon is changing at t = 1 second.</p>
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<p>We have V = (3 + t)^t (volume of the balloon)...(1) Now, we will differentiate the equation (1) Take the derivative of (3 + t)^t: dV/dt = (3 + t)^t * (ln(3 + t) + t/(3 + t)) Given t = 1, substitute this into the derivative: dV/dt = (3 + 1)^1 * (ln(3 + 1) + 1/(3 + 1)) dV/dt = 4 * (ln(4) + 1/4) Calculate the value: dV/dt = 4 * (ln(4) + 0.25) Hence, we get the rate at which the volume of the balloon is changing at t = 1 second.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of change of the balloon's volume at t = 1 second using logarithmic differentiation, substituting the given value of t, and simplifying the terms to find the result.</p>
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<p>We find the rate of change of the balloon's volume at t = 1 second using logarithmic differentiation, substituting the given value of t, and simplifying the terms to find 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 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = (x^3 + 2)^x.</p>
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<p>Derive the second derivative of the function y = (x^3 + 2)^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>The first step is to find the first derivative, dy/dx = (x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2))...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2))] Using the product rule, d²y/dx² = (x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2)) * (ln(x^3 + 2) + 3x^2/(x^3 + 2)) + (x^3 + 2)^x * (3x^2/(x^3 + 2) + 9x/(x^3 + 2)^2) Simplifying, we find the second derivative. Therefore, the second derivative of the function y = (x^3 + 2)^x is obtained by further differentiating.</p>
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<p>The first step is to find the first derivative, dy/dx = (x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2))...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2))] Using the product rule, d²y/dx² = (x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2)) * (ln(x^3 + 2) + 3x^2/(x^3 + 2)) + (x^3 + 2)^x * (3x^2/(x^3 + 2) + 9x/(x^3 + 2)^2) Simplifying, we find the second derivative. Therefore, the second derivative of the function y = (x^3 + 2)^x is obtained by further differentiating.</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, starting with the first derivative. Using the product rule, we differentiate and simplify the terms to find the second derivative.</p>
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<p>We use the step-by-step process, starting with the first derivative. Using the product rule, we differentiate and simplify the terms 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 ((2x + 1)^3) = 3(2x + 1)^2 * 2.</p>
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<p>Prove: d/dx ((2x + 1)^3) = 3(2x + 1)^2 * 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>Let’s start using the chain rule: Consider y = (2x + 1)^3 To differentiate, we use the chain rule: dy/dx = 3(2x + 1)^2 * d/dx (2x + 1) Since the derivative of (2x + 1) is 2: dy/dx = 3(2x + 1)^2 * 2 Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = (2x + 1)^3 To differentiate, we use the chain rule: dy/dx = 3(2x + 1)^2 * d/dx (2x + 1) Since the derivative of (2x + 1) is 2: dy/dx = 3(2x + 1)^2 * 2 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. Then, we replaced the inner function with its derivative to derive the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replaced the inner function with its derivative 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 ((x^2 + 1)^1/x)</p>
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<p>Solve: d/dx ((x^2 + 1)^1/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 logarithmic differentiation: Let y = (x^2 + 1)^1/x Taking the natural logarithm of both sides: ln(y) = 1/x * ln(x^2 + 1) Differentiating both sides with respect to x: 1/y * dy/dx = -1/x^2 * ln(x^2 + 1) + 1/x * (2x/(x^2 + 1)) dy/dx = y * (-1/x^2 * ln(x^2 + 1) + 2x/(x(x^2 + 1))) Substituting y = (x^2 + 1)^1/x: dy/dx = (x^2 + 1)^1/x * (-1/x^2 * ln(x^2 + 1) + 2/(x^2 + 1)) Therefore, d/dx ((x^2 + 1)^1/x) is obtained by simplifying the expression.</p>
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<p>To differentiate the function, we use logarithmic differentiation: Let y = (x^2 + 1)^1/x Taking the natural logarithm of both sides: ln(y) = 1/x * ln(x^2 + 1) Differentiating both sides with respect to x: 1/y * dy/dx = -1/x^2 * ln(x^2 + 1) + 1/x * (2x/(x^2 + 1)) dy/dx = y * (-1/x^2 * ln(x^2 + 1) + 2x/(x(x^2 + 1))) Substituting y = (x^2 + 1)^1/x: dy/dx = (x^2 + 1)^1/x * (-1/x^2 * ln(x^2 + 1) + 2/(x^2 + 1)) Therefore, d/dx ((x^2 + 1)^1/x) is obtained by simplifying the expression.</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 logarithmic differentiation. We simplify the equation to obtain the final result by applying the chain rule.</p>
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<p>In this process, we differentiate the given function using logarithmic differentiation. We simplify the equation to obtain the final result by applying the chain rule.</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 f(x)^g(x)</h2>
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<h2>FAQs on the Derivative of f(x)^g(x)</h2>
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<h3>1.Find the derivative of (x^3 + 2)^x.</h3>
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<h3>1.Find the derivative of (x^3 + 2)^x.</h3>
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<p>Using logarithmic differentiation: d/dx ((x^3 + 2)^x) = (x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2)).</p>
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<p>Using logarithmic differentiation: d/dx ((x^3 + 2)^x) = (x^3 + 2)^x * (ln(x^3 + 2) + 3x^2/(x^3 + 2)).</p>
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<h3>2.Can we use the derivative of f(x)^g(x) in real life?</h3>
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<h3>2.Can we use the derivative of f(x)^g(x) in real life?</h3>
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<p>Yes, we can use the derivative of f(x)^g(x) in real life in calculating the rate of change of any growth processes, especially in fields such as biology, economics, and physics.</p>
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<p>Yes, we can use the derivative of f(x)^g(x) in real life in calculating the rate of change of any growth processes, especially in fields such as biology, economics, and physics.</p>
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<h3>3.Is it possible to take the derivative of f(x)^g(x) if f(x) ≤ 0?</h3>
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<h3>3.Is it possible to take the derivative of f(x)^g(x) if f(x) ≤ 0?</h3>
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<p>No, f(x) must be positive for logarithmic differentiation to be valid, as the natural logarithm of a non-positive number is undefined.</p>
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<p>No, f(x) must be positive for logarithmic differentiation to be valid, as the natural logarithm of a non-positive number is undefined.</p>
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<h3>4.What rule is used to differentiate (f(x)^g(x))?</h3>
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<h3>4.What rule is used to differentiate (f(x)^g(x))?</h3>
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<p>We use logarithmic differentiation to differentiate f(x)^g(x) because it involves both power and<a>base</a>functions.</p>
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<p>We use logarithmic differentiation to differentiate f(x)^g(x) because it involves both power and<a>base</a>functions.</p>
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<h3>5.Are the derivatives of f(x)^g(x) and g(x)^f(x) the same?</h3>
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<h3>5.Are the derivatives of f(x)^g(x) and g(x)^f(x) the same?</h3>
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<p>No, they are different. The derivative of f(x)^g(x) uses logarithmic differentiation based on f(x) and g(x), whereas g(x)^f(x) would be differentiated with respect to the reverse power and base.</p>
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<p>No, they are different. The derivative of f(x)^g(x) uses logarithmic differentiation based on f(x) and g(x), whereas g(x)^f(x) would be differentiated with respect to the reverse power and base.</p>
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<h2>Important Glossaries for the Derivative of f(x)^g(x)</h2>
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<h2>Important Glossaries for the Derivative of f(x)^g(x)</h2>
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<p>Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Logarithmic Differentiation: A method used to differentiate functions of the form f(x)^g(x). Chain Rule: A fundamental rule in calculus used to differentiate composite functions. Product Rule: A rule used to differentiate products of two functions. Function: A relation or expression involving one or more variables.</p>
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<p>Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Logarithmic Differentiation: A method used to differentiate functions of the form f(x)^g(x). Chain Rule: A fundamental rule in calculus used to differentiate composite functions. Product Rule: A rule used to differentiate products of two functions. Function: A relation or expression involving one or more variables.</p>
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<p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<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>