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Magazine Chapter 7: Problem 18
Find \(f^{\prime}(x)\) if \(f(x)\) equals the given expression. $$ e^{x \ln x} $$
Short Answer
Expert verified
The derivative is \( f'(x) = e^{x \ln x} (\ln x + 1) \).
Step by step solution
01
Identify the Function Structure
Observe that the given function is \( f(x) = e^{x \ln x} \). The function is in an exponential form, \( e^{g(x)} \), with \( g(x) = x \ln x \).
02
Apply the Chain Rule
The derivative of \( f(x) = e^{g(x)} \) with respect to \( x \) can be found using the Chain Rule: \( f'(x) = e^{g(x)} \cdot g'(x) \). First, we need to find \( g'(x) \).
03
Differentiate the Inner Function \( g(x) \)
Identify \( g(x) = x \ln x \). To differentiate, apply the product rule: if \( g(x) = u(x)v(x) \) where \( u(x) = x \) and \( v(x) = \ln x \), then \( g'(x) = u'(x)v(x) + u(x)v'(x) \).
04
Apply the Product Rule
The derivative of \( u(x) = x \) is \( u'(x) = 1 \) and the derivative of \( v(x) = \ln x \) is \( v'(x) = \frac{1}{x} \). Apply the product rule: \( g'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \).
05
Calculate \( f'(x) \) Using the Chain Rule
Substitute \( g(x) = x \ln x \) and \( g'(x) = \ln x + 1 \) back into the Chain Rule: \( f'(x) = e^{x \ln x} \cdot (\ln x + 1) \).
06
Simplify the Expression
The expression for \( f'(x) \) is already simplified: \( f'(x) = e^{x \ln x} (\ln x + 1) \). This is the final expression for the derivative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. A composite function is essentially a function inside another function. When you encounter such functions, the chain rule becomes your best friend.
For a function of the form \( f(x) = e^{g(x)} \), as in our problem, the derivative involves not just the outer function \( e^{x} \), but also the inner function \( g(x) \). According to the chain rule, the derivative \( f'(x) \) is given by \( e^{g(x)} \cdot g'(x) \).
The process proceeds by calculating the derivative of the outside function \( e^{x} \), then multiplying by the derivative of the inside function \( g(x) \). This method is powerful in breaking down complex differentiations into manageable parts. Always remember:
For a function of the form \( f(x) = e^{g(x)} \), as in our problem, the derivative involves not just the outer function \( e^{x} \), but also the inner function \( g(x) \). According to the chain rule, the derivative \( f'(x) \) is given by \( e^{g(x)} \cdot g'(x) \).
The process proceeds by calculating the derivative of the outside function \( e^{x} \), then multiplying by the derivative of the inside function \( g(x) \). This method is powerful in breaking down complex differentiations into manageable parts. Always remember:
- Identify the outside and inside functions.
- Differentiate each one separately.
- Use the chain rule formula to combine your results.
Product Rule
In calculus, the product rule is a key technique used to find the derivative of products of two or more functions. If you have a function \( g(x) = u(x) \cdot v(x) \), the derivative \( g'(x) \) is obtained using the product rule:
In our specific example, \( g(x) = x \ln x \), \( u(x) = x \) and \( v(x) = \ln x \). Calculating their derivatives gives:
- \((u \, v)' = u' \, v + u \, v'\)
In our specific example, \( g(x) = x \ln x \), \( u(x) = x \) and \( v(x) = \ln x \). Calculating their derivatives gives:
- \( u'(x) = 1 \) because the derivative of \( x \) is 1.
- \( v'(x) = \frac{1}{x} \) because \( \ln x \) differentiates to \( \frac{1}{x} \).
Exponential Functions
Exponential functions are a crucial part of calculus, especially when they involve the base of the natural logarithm \( e \), like in \( e^{x \ln x} \). These functions grow rapidly and are characterized by having their rate of change proportional to their value.
The derivative of \( e^{x} \) is uniquely simple because \( \frac{d}{dx}e^{x} = e^{x} \). This property extends to \( e^{g(x)} \) by combining it with the chain rule, which gives us the product \( e^{g(x)} \cdot g'(x) \) for the derivative.
In our problem, the exponential function \( e^{x \ln x} \) increases quickly for increasing \( x \), because \( x \ln x \) itself grows. Exponential functions are integral to understanding phenomena in natural sciences, finance, and more due to their characteristic growth patterns.
The derivative of \( e^{x} \) is uniquely simple because \( \frac{d}{dx}e^{x} = e^{x} \). This property extends to \( e^{g(x)} \) by combining it with the chain rule, which gives us the product \( e^{g(x)} \cdot g'(x) \) for the derivative.
In our problem, the exponential function \( e^{x \ln x} \) increases quickly for increasing \( x \), because \( x \ln x \) itself grows. Exponential functions are integral to understanding phenomena in natural sciences, finance, and more due to their characteristic growth patterns.
