Problem 38 - Calculus Early Transcendentals

Question

Derivatives Evaluate the derivatives of the following functions. \(p(x)=x^{-\ln x}\)

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(p(x) = x^{-\ln x}\). Answer: \(\frac{dp(x)}{dx} = -x^{-\ln{x}}(2 \frac{\ln{x}}{x})\)

Step 1: Apply the natural logarithm

First, let's apply the natural logarithm to the function and use the property \(y = e^{\ln y}\) \(\ln{p(x)} = \ln{\left(x^{-\ln x}\right)}\), so \(p(x) = e^{\ln{\left(x^{-\ln x}\right)}}\). Now, we can also simplify the exponent using logarithm properties \(\ln{p(x)} = -\ln{x} \cdot \ln{x}\).

Step 2: Take the derivative of the logarithm

Differentiate both sides with respect to x using the chain rule and the derivative of the natural logarithm function. \(\frac{d}{dx} \ln{p(x)} = \frac{d}{dx}(-\ln{x} \cdot \ln{x})\) As we're deriving \(p(x)\), we should derive it with respect to \(x\). Hence we use the chain rule: \(\frac{1}{p(x)} \cdot \frac{dp(x)}{dx} = -(\ln{x} \cdot \frac{d \ln{x}}{dx} + \frac{d \ln{x}}{dx} \cdot \ln{x})\) Now, differentiate \(\ln{x}\) with respect to x: \(\frac{d \ln{x}}{dx} = \frac{1}{x}\) Replace the derivative in the equation: \(\frac{1}{p(x)} \cdot \frac{dp(x)}{dx} = -(\ln{x} \cdot \frac{1}{x} + \frac{1}{x} \cdot \ln{x})\)

Step 3: Solve for the derivative

Multiply both sides by \(p(x)\) to isolate the derivative. \(\frac{dp(x)}{dx} = -p(x)(2 \frac{\ln{x}}{x})\) Recall that \(p(x) = x^{-\ln x}\), so the derivative, \(\frac{dp(x)}{dx} = -x^{-\ln{x}}(2 \frac{\ln{x}}{x})\) Now we have found the derivative of the given function.

Key Concepts

Chain Rule

Understanding how to apply the chain rule is crucial, especially when differentiating composite functions. The chain rule allows us to find the derivative of a function that is comprised of other functions. This is defined as: if we have a function that can be described as a composite, such as \( f(g(x)) \), the derivative \( f'(x) \) can be calculated using this rule. The chain rule states:

• Differentiate the outer function \( f \) with respect to the inner function \( g \).
• Multiply this result by the derivative of the inner function \( g \) with respect to \( x \).

In mathematical terms, this is expressed as:\[(f(g(x)))' = f'(g(x)) \, g'(x)\]In the provided solution, the chain rule is applied in step 2 when deriving \( \ln{p(x)} \). This involves differentiating \( -\ln{x} \, \ln{x} \), which is a product of logarithmic functions inside the logarithm. By using the chain rule, we are able to effectively manage the complexity of differentiating composite expressions.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique, especially when dealing with functions that have variables in both the base and the exponent. This method is particularly useful for simplifying the differentiation of such functions. Here's how it works:

• Take the natural logarithm of both sides of an equation \( y = f(x) \), which often helps simplify complex expressions.
• Differentiate each side with respect to \( x \), using the properties of logarithms to simplify where possible.
• Solve for the derivative \( \frac{dy}{dx} \).

In our example, the original function \( x^{-\ln x} \) was transformed into a more manageable form by taking its natural logarithm.This simplifies our differentiation, turning multiplication and division into addition and subtraction. We transform \( \ln{p(x)} = -\ln{x} \, \ln{x} \), making it easier to apply differentiation rules including the chain rule, and eventually isolating \( \frac{dp(x)}{dx} \).

Natural Logarithm Properties

The properties of the natural logarithm, denoted as \( \ln \), are pivotal in simplifying expressions before differentiation. By utilizing these properties, expressions involving exponents and roots can often be transformed into easier forms. Some key properties include:

• \( \ln(a \cdot b) = \ln a + \ln b \) - allows us to split products into sums.
• \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \) - helps in differentiating quotients by turning into differences.
• \( \ln(a^b) = b \cdot \ln a \) - is particularly useful for handling exponents, as seen in our example where \( \ln(x^{b}) \) reduces to \( b \cdot \ln x \).

In the solution, we use these properties to transform \( \ln\left(x^{-\ln x}\right)\) into \( -\ln{x} \cdot \ln{x} \). This approach simplifies the differentiation process significantly since it converts the complexity of exponents into a simple multiplication operation that is straightforward to differentiate.