Problem 65
Question
Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=(\ln x)^{x} $$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{dy}{dx} = (\ln x)^x \left( \ln(\ln x) + \frac{1}{\ln x} \right) \).
1Step 1: Apply Logarithmic Differentiation
To begin, recognize that the function is of the form \( f(x) = (\ln x)^x \), which makes it suitable for logarithmic differentiation. Take the natural logarithm of both sides, which transforms the equation to \( \ln y = \ln((\ln x)^x) = x \ln(\ln x) \).
2Step 2: Differentiate Both Sides Implicitly
Differentiate both sides with respect to \( x \). The left side differentiation gives \( \frac{1}{y} \frac{dy}{dx} \), and for the right side, use the product rule: \( \frac{d}{dx}(x \ln(\ln x)) = 1 \cdot \ln(\ln x) + x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} = \ln(\ln x) + \frac{1}{\ln x} \).
3Step 3: Solve for the Derivative
Reorganize the derivative expression to solve for \( \frac{dy}{dx} \): \( \frac{1}{y} \frac{dy}{dx} = \ln(\ln x) + \frac{1}{\ln x} \). Then, multiply both sides by \( y = (\ln x)^x \) to clear the fraction: \( \frac{dy}{dx} = (\ln x)^x \left( \ln(\ln x) + \frac{1}{\ln x} \right) \).
Key Concepts
Understanding Logarithmic Differentiation in CalculusExploring Derivatives and their ImportanceImplicit Differentiation: A Deeper Dive
Understanding Logarithmic Differentiation in Calculus
Logarithmic differentiation is a powerful technique in calculus, especially useful for differentiating functions that involve variables raised to the power of other variables. This method employs the natural logarithm to transform a complex expression into a more manageable one.
When applied to the function \( f(x) = (\ln x)^x \), the first step involves taking the natural logarithm of both sides, turning the product of exponents into a simpler form involving sums. This drastically simplifies the differentiation task, allowing us to handle complicated functions as seen in our exercise example.
Logarithmic differentiation is not only handy for functions with variables as exponents but also comes in clutch when dealing with products and quotients of functions as well. It breaks down expressions, making the calculus operation straightforward and manageable. However, it should be noted that it requires the original function to be positive, which is usually the case for functions involving logarithms.
When applied to the function \( f(x) = (\ln x)^x \), the first step involves taking the natural logarithm of both sides, turning the product of exponents into a simpler form involving sums. This drastically simplifies the differentiation task, allowing us to handle complicated functions as seen in our exercise example.
Logarithmic differentiation is not only handy for functions with variables as exponents but also comes in clutch when dealing with products and quotients of functions as well. It breaks down expressions, making the calculus operation straightforward and manageable. However, it should be noted that it requires the original function to be positive, which is usually the case for functions involving logarithms.
Exploring Derivatives and their Importance
Derivatives are a core concept in calculus, representing the rate of change of a function concerning one of its variables. In simpler terms, differentiating a function tells you how much the function's output changes when you slightly tweak the input.
In our exercise, finding the derivative of \( f(x) = (\ln x)^x \) reveals the instantaneous rate of change of this function as \( x \) varies. The steps show how applying logarithmic differentiation and implicit differentiation simplify this process.
Understanding derivatives is crucial because they serve as the foundation for many concepts in calculus and physics. Here are some key reasons they are invaluable:
In our exercise, finding the derivative of \( f(x) = (\ln x)^x \) reveals the instantaneous rate of change of this function as \( x \) varies. The steps show how applying logarithmic differentiation and implicit differentiation simplify this process.
Understanding derivatives is crucial because they serve as the foundation for many concepts in calculus and physics. Here are some key reasons they are invaluable:
- Optimization: Identifying maximum and minimum values of functions.
- Describing Motion: Used in physics to find velocity and acceleration.
- Predicting Changes: Helping in economics to study marginal costs and revenues.
Implicit Differentiation: A Deeper Dive
Implicit differentiation is a technique used when dealing with equations where \( y \) is not isolated. Instead of solving for \( y \) explicitly and differentiating, you differentiate both sides concerning \( x \), treating \( y \) as a function of \( x \). This is crucial when it's hard or impossible to express \( y \) strictly as a function of \( x \).
In the context of logarithmic differentiation, once we've applied the natural log to \( f(x) = (\ln x)^x \), implicit differentiation helps in dealing with the transformed equation \( \ln y = x \ln(\ln x) \). By differentiating both sides, respecting the chain rule, it provides a path to find the derivative \( \frac{dy}{dx} \) of the starting function.
Key points about implicit differentiation include:
In the context of logarithmic differentiation, once we've applied the natural log to \( f(x) = (\ln x)^x \), implicit differentiation helps in dealing with the transformed equation \( \ln y = x \ln(\ln x) \). By differentiating both sides, respecting the chain rule, it provides a path to find the derivative \( \frac{dy}{dx} \) of the starting function.
Key points about implicit differentiation include:
- It's useful for differentiating equations where variables are intermingled.
- By considering \( y \) as a function of \( x \), it accounts for how both variables change.
- It leverages the derivative directly within an equation without needing explicit functional relationships.
Other exercises in this chapter
Problem 65
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