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Magazine Chapter 6: Problem 8
Prove that the function \(f(x)=x^{x}, x \in(0, \infty)\), is differentiable on its domain, and find its derivative.
Short Answer
Expert verified
The derivative of \( f(x) = x^x \) is \( f'(x) = x^x (\ln x + 1) \).
Step by step solution
01
Understanding the Function
The function given is \( f(x) = x^x \). It is defined for \( x > 0 \) because the base \( x \) must be positive to consider the power \( x \). We are required to prove that this function is differentiable over its domain, which is \((0, \infty)\).
02
Rewrite the Function
To differentiate the function, it's easier to work with a natural logarithm. Rewrite the function as \( f(x) = e^{x \ln x} \). This is accomplished by using the property of exponents: \( x^x = e^{\ln(x^x)} = e^{x\ln x} \).
03
Differentiate Using the Chain Rule
Differentiate \( f(x) = e^{x \ln x} \) with respect to \( x \) using the chain rule. Let \( u = x \ln x \), so \( f(x) = e^u \). The derivative of \( e^u \) with respect to \( x \) is \( e^u \cdot \frac{du}{dx} \).
04
Find the Derivative of \( u = x \ln x \)
Differentiate \( u = x \ln x \) with respect to \( x \). Use the product rule, where \( u = x \) and \( v = \ln x \), giving \( \frac{du}{dx} = x'v + xv' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \).
05
Combine Derivatives
Now, combine the derivatives from previous steps: \( \frac{d}{dx}[e^{x \ln x}] = e^{x \ln x} \cdot (\ln x + 1) \). Since \( e^{x\ln x} = x^x \), the derivative \( f'(x) = x^x (\ln x + 1) \).
06
Verify Differentiability
The derivative \( f'(x) = x^{x}(\ln x + 1) \) is defined for all \( x > 0 \), thus proving that \( f(x) \) is differentiable over the interval \((0, \infty)\).
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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 principle in calculus used to find the derivative of composite functions. Imagine you have a function nested inside another function, like a Russian doll.
For instance, consider the function \( f(x) = e^{x \ln x} \). This is a composite function where \( u = x \ln x \) is inside the exponent of \( e^u \). To differentiate \( f(x) \), you employ the Chain Rule: differentiate the outer function, \( e^u \), and then multiply it by the derivative of the inner function, \( u \).
This gives us:
For instance, consider the function \( f(x) = e^{x \ln x} \). This is a composite function where \( u = x \ln x \) is inside the exponent of \( e^u \). To differentiate \( f(x) \), you employ the Chain Rule: differentiate the outer function, \( e^u \), and then multiply it by the derivative of the inner function, \( u \).
This gives us:
- Differentiate \( e^u \) to get \( e^u \).
- Then, multiply by \( \frac{du}{dx} \), the derivative of \( x \ln x \).
Product Rule
The Product Rule is another essential tool in differentiation, ideal for handling functions that are products of two or more separate functions. When you're tasked with finding the derivative of a product like \( u(x) \, v(x) \), the Product Rule breaks it down into manageable parts.
The general formula is \( \frac{d}{dx}[u \, v] = u' \, v + u \, v' \), where \( u' \) and \( v' \) denote the derivatives of \( u \) and \( v \) respectively.
For the function \( u = x \ln x \) we used the Product Rule, treating \( u \) as \( x \) and \( v \) as \( \ln x \). Here's how it works:
The general formula is \( \frac{d}{dx}[u \, v] = u' \, v + u \, v' \), where \( u' \) and \( v' \) denote the derivatives of \( u \) and \( v \) respectively.
For the function \( u = x \ln x \) we used the Product Rule, treating \( u \) as \( x \) and \( v \) as \( \ln x \). Here's how it works:
- First, differentiate \( x \), giving \( 1 \).
- Next, differentiate \( \ln x \), giving \( \frac{1}{x} \).
- Apply the Product Rule as: \( 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \).
Derivative of Exponential Functions
Differentiating exponential functions might seem daunting, but it's straightforward with the right approach.
In calculus, the exponential function \( e^{u(x)} \) has a specific differentiation rule: the derivative is the product of the function \( e^{u(x)} \) and the derivative of its exponent \( u(x) \). The result is \( \frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot \frac{du}{dx} \).
Let's take the function \( f(x) = e^{x \ln x} \):
In calculus, the exponential function \( e^{u(x)} \) has a specific differentiation rule: the derivative is the product of the function \( e^{u(x)} \) and the derivative of its exponent \( u(x) \). The result is \( \frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot \frac{du}{dx} \).
Let's take the function \( f(x) = e^{x \ln x} \):
- The derivative is \( e^{x \ln x} \) times the derivative of \( x \ln x \), which is \( \ln x + 1 \).
- Therefore, \( \frac{d}{dx}[e^{x \ln x}] = e^{x \ln x} \cdot (\ln x + 1) \).
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is the logarithm to the base \( e \), an irrational constant approximately equal to 2.71828. It plays a critical role, particularly in calculus, because of its unique properties.
One important feature of the natural logarithm is its simple differentiation rule. The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \). This makes it very handy when working with functions involving \( \ln x \).
Returning to our example:
One important feature of the natural logarithm is its simple differentiation rule. The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \). This makes it very handy when working with functions involving \( \ln x \).
Returning to our example:
- In differentiating \( u = x \ln x \), we see \( \ln x \) and instantly know its derivative, simplifying the process greatly.
- This property was used in conjunction with the Product Rule to find that \( \frac{du}{dx} = \ln x + 1 \).
