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Magazine Chapter 4: Problem 69
Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=x^{\ln x} $$
Short Answer
Expert verified
The derivative of \( f(x) = x^{\ln x} \) is \( f'(x) = 2x^{\ln x - 1} \ln x \).
Step by step solution
01
Take the Natural Logarithm of Both Sides
Start by taking the natural logarithm of both sides of the equation to simplify the differentiation process. This gives us \( \ln f(x) = \ln(x^{\ln x}) \).
02
Apply Logarithm Power Rule
Use the logarithm power rule \( \ln(a^b) = b\ln a \). Thus, the equation becomes \( \ln f(x) = (\ln x) \cdot (\ln x) = (\ln x)^2 \).
03
Differentiate Implicitly with Respect to x
Differentiate both sides with respect to \( x \). The left side, \( \ln f(x) \), differentiates to \( \frac{1}{f(x)} \cdot f'(x) \) using the chain rule. The right side, \( (\ln x)^2 \), uses the chain rule for differentiation, resulting in \( 2 \ln x \cdot \frac{1}{x} \).
04
Solve for \( f'(x) \)
Equate the derivatives obtained: \( \frac{f'(x)}{f(x)} = \frac{2 \ln x}{x} \). Solve for \( f'(x) \) by multiplying both sides by \( f(x) \), giving us \( f'(x) = f(x) \cdot \frac{2 \ln x}{x} \).
05
Substitute \( f(x) \) Back into the Equation
Substitute \( f(x) = x^{\ln x} \) back in to get: \( f'(x) = x^{\ln x} \cdot \frac{2 \ln x}{x} = 2x^{\ln x - 1} \cdot \ln x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is an essential concept in calculus and related fields. Its base is the constant \( e \), approximately 2.718, and it's commonly used to simplify the differentiation of complex functions.
Natural logarithms make dealing with exponential terms easier, especially when higher powers and roots are involved. For instance, when we have powers like \( x^{\ln x} \), taking the logarithm helps to transform multiplicative processes into additive ones.
Natural logarithms make dealing with exponential terms easier, especially when higher powers and roots are involved. For instance, when we have powers like \( x^{\ln x} \), taking the logarithm helps to transform multiplicative processes into additive ones.
- This transformation leverages the fact that the logarithm of a power, \( \ln(a^b) = b \cdot \ln a \), allows us to differentiate more easily.
- Most critically, using natural logarithms helps in handling composite functions, as it can simplify the operations needed to find derivatives.
Implicit Differentiation
Implicit differentiation is a technique often used when we have functions intermingled together, sometimes not neatly defined as \( y = f(x) \). Instead, functions are given in relationships like \( \ln f(x) = (\ln x)^2 \).
This method helps when functions are not easily separated. For this particular problem, the function \( f(x) \) itself wasn't initially isolated, so implicit differentiation is required.
This method helps when functions are not easily separated. For this particular problem, the function \( f(x) \) itself wasn't initially isolated, so implicit differentiation is required.
- The core idea is to differentiate both sides of the equation with respect to \( x \), even if \( y \) or \( f(x) \) is not explicitly stated as \( y \) in terms of \( x \).
- In the process, we assume \( f(x) \) as a function of \( x \), denoting its derivative as \( f'(x) \).
- This assumption leads us to use the chain rule to manage composite functions, linking relative rates of change.
Chain Rule
The chain rule is another fundamental concept in differentiation, tailor-made for composite functions. It allows you to differentiate a function based on its composition, by breaking it down into smaller, easier-to-manage parts.
The structure of composite functions, such as \( \ln(f(x)) \), necessitates the use of the chain rule. This is because the derivative of \( \ln(f(x)) \) involves differentiating both the outer function \( \ln \,\) and the inner function \( f(x) \).
The structure of composite functions, such as \( \ln(f(x)) \), necessitates the use of the chain rule. This is because the derivative of \( \ln(f(x)) \) involves differentiating both the outer function \( \ln \,\) and the inner function \( f(x) \).
- The rule states that the derivative of \( h(x) = f(g(x)) \) is \( h'(x) = f'(g(x)) \cdot g'(x) \).
- For the problem at hand, applying the chain rule to \( \ln f(x) \) yielded \( \frac{1}{f(x)} \cdot f'(x) \).
- Similarly, for \( (\ln x)^2 \), we differentiated using the power and chain rules together to arrive at \( 2 \ln x \cdot \frac{1}{x} \).
Power Rule
The power rule provides a straightforward guideline for finding the derivative of polynomial functions. It states: If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).
In the exercise, where \( y = (\ln x)^2 \), the power rule plays an essential role along with the chain rule.
In the exercise, where \( y = (\ln x)^2 \), the power rule plays an essential role along with the chain rule.
- The application of this rule simplifies the process of finding derivatives with polynomial forms, even when the base of the polynomial involves a logarithm or other complex expressions.
- When combining the power rule with the natural logarithm, such as in \( (\ln x)^2 \), it's essential to remember the additional chain rule component that results from the base being a function, not just \( x \).
- Thus, differentiating \( (\ln x)^2 \) involves initially applying the power rule to get \( 2(\ln x) \), paired with the inner derivative \( \frac{1}{x} \) from the chain rule.
