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Last updated on July 19th, 2025

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Derivative of f(x)^g(x)

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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.

Derivative of f(x)^g(x) for US Students
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What is the Derivative of f(x)^g(x)?

We now understand the derivative of f(x)^g(x). It is commonly represented as d/dx (f(x)^g(x)) and can be derived using logarithmic differentiation. The function 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.

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Derivative of f(x)^g(x) Formula

The derivative of f(x)^g(x) can be obtained using logarithmic differentiation. The formula 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.

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Proofs of the Derivative of f(x)^g(x)

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 exponent 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 product 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))

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Higher-Order Derivatives of f(x)^g(x)

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 rate 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.

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Special Cases:

When f(x) = x^0, the derivative is zero because any number raised to the power of zero is 1, and the derivative of a constant is zero. When g(x) = 0, the derivative of f(x)^0 = 1, which is also zero, as it is a constant function.

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Common Mistakes and How to Avoid Them in Derivatives of f(x)^g(x)

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:

Mistake 1

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Not applying logarithmic differentiation

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Students may forget to apply logarithmic differentiation, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result. Ensure that each step is written in order. It might seem tedious, but it is crucial to avoid errors in the process.

Mistake 2

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Ignoring the domain constraints

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They might not remember that f(x) must be positive for the logarithmic differentiation to be valid. Keep in mind that you should consider the domain of the function you differentiate. This will help understand that the function is not continuous at certain points if f(x) ≤ 0.

Mistake 3

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Incorrect use of chain rule

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While differentiating functions such as (f(x)^g(x)), students misapply the chain rule. For example: Incorrect differentiation: d/dx (f(x)^g(x)) = g(x)f(x)^(g(x)-1)f'(x). To avoid this mistake, apply the chain rule correctly, considering the inner and outer functions separately. Always check for errors in the calculation and ensure it is properly simplified.

Mistake 4

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Not considering the constant functions

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Students sometimes overlook the fact that a function raised to the power of zero is a constant. For example, they incorrectly write d/dx (f(x)^0) = 0 when it should be zero without differentiation. Recognize constant functions and understand that their derivative is zero.

Mistake 5

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Not applying the product rule

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Students often forget to use the product rule when differentiating complex products within f(x)^g(x). This happens when the derivative of the product is not considered. To fix this error, identify each function component and apply the product rule as needed.

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Examples Using the Derivative of f(x)^g(x)

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Problem 1

Calculate the derivative of ((x^2 + 1)^x).

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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)).

Explanation

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.

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Problem 2

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.

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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.

Explanation

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.

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Problem 3

Derive the second derivative of the function y = (x^3 + 2)^x.

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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.

Explanation

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.

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Problem 4

Prove: d/dx ((2x + 1)^3) = 3(2x + 1)^2 * 2.

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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.

Explanation

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.

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Problem 5

Solve: d/dx ((x^2 + 1)^1/x)

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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.

Explanation

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.

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FAQs on the Derivative of f(x)^g(x)

1.Find the derivative of (x^3 + 2)^x.

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2.Can we use the derivative of f(x)^g(x) in real life?

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3.Is it possible to take the derivative of f(x)^g(x) if f(x) ≤ 0?

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4.What rule is used to differentiate (f(x)^g(x))?

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5.Are the derivatives of f(x)^g(x) and g(x)^f(x) the same?

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Important Glossaries for the Derivative of f(x)^g(x)

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.

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Jaskaran Singh Saluja

About the Author

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.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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