Problem 71 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=x^{1 / x} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{x^{1/x} (1 - \ln x)}{x^2} \).

Step 1: Take Natural Logarithm on Both Sides

To simplify the differentiation process, take the natural logarithm of both sides of the function. We use: \( y = f(x) = x^{1/x} \). Thus, \( \ln y = \ln(x^{1/x}) \). By the properties of logarithms, this becomes \( \ln y = \frac{1}{x} \ln x \).

Step 2: Differentiate Implicitly with Respect to x

Differentiate both sides of the equation \( \ln y = \frac{1}{x} \ln x \) with respect to \( x \). For the left side, use the chain rule: \( \frac{1}{y} \frac{dy}{dx} \). For the right side, apply the product rule to \( \frac{1}{x} \ln x \), yielding: \( \frac{1}{x} \cdot \frac{1}{x} + \ln x \cdot \left( -\frac{1}{x^2} \right) \).

Step 3: Simplify the Derivative

Simplify the right side of the equation: \( \frac{1}{x^2} - \frac{\ln x}{x^2} \). Thus the equation becomes \( \frac{1}{y} \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \). Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = y \cdot \frac{1 - \ln x}{x^2} \).

Step 4: Substitute Back the Function for y

Recall that \( y = x^{1/x} \). Substitute \( y \) in the expression for \( \frac{dy}{dx} \), giving \( \frac{dy}{dx} = x^{1/x} \cdot \frac{1 - \ln x}{x^2} \).

Step 5: Finalize the Derivative Expression

Simplify the final expression further if necessary. The first derivative of \( f(x) = x^{1/x} \) is therefore \( f'(x) = \frac{x^{1/x} (1 - \ln x)}{x^2} \).

Key Concepts

First Derivative

Differentiating a function means finding its rate of change or how the function's output changes with a tiny change in the input. The first derivative, typically expressed as \( f'(x) \), tells us exactly this rate of change at any point \( x \) on the curve of the function. When dealing with complex functions, differentiating directly can be challenging. For the function \( f(x) = x^{1/x} \), a direct differentiation approach is not straightforward due to the nature of the exponent variable. This is where logarithmic differentiation comes into play, providing a simplified method by utilizing the property of natural logarithms. By taking the natural logarithm of both sides of our function \( y = x^{1/x} \), we transform the differentiation process into a more manageable task. Recall, the power rule, chain rule, and properties of logarithms are tools that help convert multiplication and exponentiation into addition or subtraction within a logarithmic context, easing the differentiation process. Therefore, the initial step involves rewriting the function as \( \ln y = \frac{1}{x} \ln x \), which sets the stage for the utilization of implicit differentiation.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given explicitly as \( y = f(x) \), but rather in a relational context between \( x \) and \( y \). In this exercise, after applying the natural logarithm, our expression becomes \( \ln y = \frac{1}{x} \ln x \). Although implicitly shown, it allows us to differentiate both sides with respect to \( x \) conveniently.On the left side, we apply the chain rule to get \( \frac{1}{y} \frac{dy}{dx} \). For the right side, it's crucial to recognize that \( \frac{1}{x} \ln x \) is a product of two separate functions: \( \frac{1}{x} \) and \( \ln x \). Here, the product rule is required, as it helps handle the multiplication of two functions.After applying implicit differentiation correctly, you'll find yourself with an equation where \( \frac{dy}{dx} \), or the derivative, is expressed in terms of both \( x \) and \( y \), facilitating the next step of solving this for \( \frac{dy}{dx} \). This culmination is a gateway to find the first derivative of the original function by substituting back \( y \) as \( x^{1/x} \).

Product Rule

The product rule is a fundamental tool in calculus that helps differentiate expressions where two functions are multiplied together. It states that for functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is calculated as \( u'(x)v(x) + u(x)v'(x) \).When dealing with \( \frac{1}{x} \ln x \) on the right side of \( \ln y = \frac{1}{x} \ln x \), we recognize it as a product of the functions \( u(x) = \frac{1}{x} \) and \( v(x) = \ln x \). Differentiating \( u(x) \) results in \( -\frac{1}{x^2} \), and differentiating \( v(x) \) yields \( \frac{1}{x} \).Applying the product rule:

• Differentiate the first component \( \frac{1}{x} \) as \( -\frac{1}{x^2} \).
• Differentiate the second component \( \ln x \) as \( \frac{1}{x} \).
• Combine the results: \( \frac{1}{x} \cdot \frac{1}{x} + \ln x \cdot \left( -\frac{1}{x^2} \right) \).

This results in an expression \( \frac{1}{x^2} - \frac{\ln x}{x^2} \) that is critical for simplifying the calculation of \( \frac{dy}{dx} \). Embracing the product rule in this differentiation process shows how interconnected techniques in calculus can break down complex derivations into simpler, clear steps.