Problem 72 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=x^{3 / x} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = x^{3/x} \cdot \left( \frac{3 - 3\cdot \text{ln}(x)}{x^2} \right) \).

Step 1: Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm of both sides of the equation to simplify the function raised to a power. \( ext{Let } y = f(x) = x^{3/x} \ ext{Take the natural log: } \ ext{ln}(y) = ext{ln}(x^{3/x}) \)

Step 2: Use Logarithm Power Rule

Apply the power rule of logarithms, which states that \( \text{ln}(a^b) = b \cdot \text{ln}(a) \), to the right side of the equation.\( \text{ln}(y) = \frac{3}{x} \cdot \text{ln}(x) \)

Step 3: Differentiate Implicitly with Respect to x

Differentiate both sides with respect to \( x \), remembering that \( y \) is a function of \( x \).\( \frac{d}{dx}[\text{ln}(y)] = \frac{d}{dx} \left( \frac{3}{x} \cdot \text{ln}(x) \right) \)The derivative of \( \text{ln}(y) \) is \( \frac{1}{y} \cdot \frac{dy}{dx} \) and the right side requires the product rule and chain rule.

Step 4: Apply the Product Rule on Right Side

Apply the product rule: If \( u = \frac{3}{x} \) and \( v = \text{ln}(x) \), then \( \frac{d}{dx}(uv) = u'v + uv' \).\( u = \frac{3}{x} \Rightarrow u' = -\frac{3}{x^2} \) \( v = \text{ln}(x) \Rightarrow v' = \frac{1}{x} \)\( \frac{d}{dx} \left( \frac{3}{x} \cdot \text{ln}(x) \right) = -\frac{3}{x^2} \cdot \text{ln}(x) + \frac{3}{x} \cdot \frac{1}{x} \)

Step 5: Simplify the Derivative

Combine and simplify the terms from Step 4.\( \frac{d}{dx} \left( \frac{3}{x} \cdot \text{ln}(x) \right) = -\frac{3\cdot \text{ln}(x)}{x^2} + \frac{3}{x^2} \)

Step 6: Solve for the Derivative \( \frac{dy}{dx} \)

Use the result from Step 3 and continue solving for \( \frac{dy}{dx} \).Plug the expression into Step 3's derivative: \( \frac{1}{y} \cdot \frac{dy}{dx} = -\frac{3\cdot \text{ln}(x)}{x^2} + \frac{3}{x^2} \ \Rightarrow \frac{dy}{dx} = y \left( -\frac{3\cdot \text{ln}(x)}{x^2} + \frac{3}{x^2} \right) \) Substitute back \( y = x^{3/x} \): \( \frac{dy}{dx} = x^{3/x} \left( -\frac{3\cdot \text{ln}(x)}{x^2} + \frac{3}{x^2} \right) \)

Step 7: Finalize the Expression

The derivative \( \frac{dy}{dx} \) is now: \( \frac{dy}{dx} = x^{3/x} \cdot \left( \frac{3 - 3\cdot \text{ln}(x)}{x^2} \right) \). Simply be sure to express this result neatly and in the simplest form possible.

Key Concepts

Derivative

In calculus, the concept of a derivative is fundamental. Derivatives tell us how a function changes as its input changes. They are like a mathematical tool for finding rates of change. For example, if you have a function that describes the distance a car travels, the derivative of this function will tell you the car's speed at any moment. When we say we are finding the first derivative, we are talking about finding the primary rate at which the function's value changes with respect to its input. In the given exercise, you're tasked with finding the first derivative of the function \( f(x) = x^{3/x} \). To do this properly, you can utilize logarithmic differentiation, which is especially useful when dealing with functions where both the base and the exponent involve variables, such as \( x^{3/x} \). This method simplifies the differentiation process considerably.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a special type of logarithm with the base \( e \), where \( e \) is an irrational and transcendental constant approximately equal to 2.71828. The natural logarithm is especially useful in calculus due to its simple derivative: the derivative of \( \ln(x) \) is \( 1/x \). This property simplifies the process of differentiation, as shown in the exercise where \( \ln(y) = \ln(x^{3/x}) \). By using the natural logarithm, we convert the exponent into a product \( \frac{3}{x} \ln(x) \), making it easier to differentiate. Hence, the natural logarithm helps us break down complex functions into more manageable forms during the differentiation process.

Product Rule

The product rule is a crucial differentiation technique used when you're dealing with the multiplication of two functions. Suppose you have two functions, \( u(x) \) and \( v(x) \). The product rule tells us that the derivative of their product is given by \( u'v + uv' \), where \( u' \) and \( v' \) are derivatives of \( u \) and \( v \), respectively. In the context of the problem \( f(x) = x^{3/x} \), after applying the logarithmic and power rules, the natural log transform results in the function \( \ln(y) = \frac{3}{x} \cdot \ln(x) \). Here, \( u = \frac{3}{x} \) and \( v = \ln(x) \). You apply the product rule to find the derivative of this product, which leads to \( u'v + uv' \), thus allowing us to solve the problem effectively and find \( \frac{d}{dx} \left( \frac{3}{x} \cdot \ln(x) \right) \).

Implicit Differentiation

Implicit differentiation is a technique used when it is difficult or impossible to solve for one variable in terms of another explicitly. Essentially, it allows us to find the derivative of a variable with respect to another when the relationship between them is given implicitly rather than explicitly. In the exercise where \( y = x^{3/x} \), implicit differentiation comes into play after taking the natural logarithm of both sides. We do not solve for \( y \) explicitly in terms of \( x \) but instead differentiate directly with respect to \( x \). By recognizing \( y \) as a function of \( x \), we use implicit differentiation to find \( \frac{dy}{dx} \). This involves differentiating \( \ln(y) = \frac{3}{x} \ln(x) \) implicitly by applying the chain rule, resulting in \( \frac{1}{y} \cdot \frac{dy}{dx} \). Thus, implicit differentiation enables us to maneuver through complex relationships and find derivatives even when functions are intertwined in complicated ways.