Problem 323 - Calculus Volume 1

Question

For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$y=x^{-1 / x}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = x^{-1/x} \left( \frac{1}{x^2} \ln x - \frac{1}{x^2} \right) \)

Step 1: Rewrite the Expression

The function given is \( y = x^{-1/x} \). Start by rewriting it in terms of logarithms to facilitate differentiation. Take the natural logarithm of both sides: \( \ln y = \ln(x^{-1/x}) \). Apply the logarithmic identity, \( \ln(a^b) = b \ln a \), to simplify to \( \ln y = -\frac{1}{x} \ln x \).

Step 2: Differentiate Both Sides

Differentiate both sides of the equation with respect to \( x \). The left side becomes \( \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \). The right side requires the product rule and chain rule; let \( u = -\frac{1}{x} \) and \( v = \ln x \), then \( \frac{d}{dx}(uv) = u'v + uv' \). Calculate derivatives: \( u' = \frac{1}{x^2} \) and \( v' = \frac{1}{x} \). So, \( \frac{d}{dx}(-\frac{1}{x} \ln x) = \frac{1}{x^2} \ln x - \frac{1}{x^2} \).

Step 3: Solve for \( \frac{dy}{dx} \)

Substitute the differentiated expressions into \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2} \ln x - \frac{1}{x^2} \). Solve for \( \frac{dy}{dx} \) by multiplying through by \( y \), giving \( \frac{dy}{dx} = y \left( \frac{1}{x^2} \ln x - \frac{1}{x^2} \right) \). Substitute back \( y = x^{-1/x} \) to express \( \frac{dy}{dx} \) explicitly in terms of \( x \): \( \frac{dy}{dx} = x^{-1/x} \left( \frac{1}{x^2} \ln x - \frac{1}{x^2} \right) \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used when differentiating functions that are difficult to handle using straightforward methods. It is particularly useful when dealing with products, quotients, or powers of functions. The main idea is to take the natural logarithm of both sides of the equation to simplify the differentiation process.
Here’s how it works:

• Start by taking the natural logarithm on both sides of the function you want to differentiate. For example, if you have a function like \( y = x^{-1/x} \), begin by stating \( \ln y = \ln (x^{-1/x}) \).
• Utilize properties of logarithms, such as \( \ln(a^b) = b \ln(a) \), to make the expression easier to work with. In this exercise, this becomes \( \ln y = -\frac{1}{x} \ln x \).
• Differentiate both sides concerning \( x \). Be careful to apply additional differentiation rules such as the product and chain rule, as needed.

By reducing complexity with logarithms, you gain simplicity in managing complex derivatives, enhancing both accuracy and efficiency in finding solutions.

Product Rule

The product rule is a key differentiation tool needed when differentiating products of functions. When two functions are multiplied together, their derivative isn't simply the product of their individual derivatives. Instead, use the product rule.
The product rule states that if \( u \) and \( v \) are functions of \( x \), then the derivative of their product is:
\[ \frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v' \]
In the context of our original problem, the expression \( -\frac{1}{x} \ln x \) can be seen as the product of two functions: \( u = -\frac{1}{x} \) and \( v = \ln x \).
Using the product rule here, we get:

• Determine \( u' \), the derivative of \( u \), which is \( \frac{1}{x^2} \).
• Determine \( v' \), the derivative of \( v \), which is \( \frac{1}{x} \).
• Then, applying the product rule: \( u'v + uv' = \frac{1}{x^2} \ln x - \frac{1}{x^2} \).

This approach makes the differentiation process both systematic and manageable, especially for products involving logarithms.

Chain Rule

The chain rule is another essential differentiation technique utilized for functions within functions, often referred to as composite functions. It states that the derivative of a composite function \( f(g(x)) \) can be found by multiplying the derivative of \( f \) with respect to \( g \) and the derivative of \( g \) with respect to \( x \):
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
In our original problem, although the chain rule isn't explicitly used as a standalone technique, it often works behind the scenes in combination with other rules like the product rule. For example, consider \( v = \ln x \), differentiating this requires recognizing it as a simple composite function where \( f(x) = \ln(x) \) and \( g(x) = x \).
Using the chain rule is akin to peeling an onion:

• Focus on the outer function, finding its derivative. Here, \( f'(g(x)) \) becomes \( \frac{1}{x} \), as the derivative of \( \ln(x) \) is \( \frac{1}{x} \).
• Multiply it by the "inner" function's derivative, which is 1 in this simple case, keeping the differentiation smooth and efficient.

Understanding and applying the chain rule allows us to navigate complex functions with nested components, enhancing problem-solving versatility.