Problem 58 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$ y=\sqrt{x}^{x} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{x^{x/2}}{2} (\ln x + 1) \).

Step 1: Express the function using logarithms

First, recognize that the function can be rewritten for easier differentiation. Using properties of logarithms, express \( y = \sqrt{x}^x \) as \( y = x^{x/2} \). Taking the natural logarithm of both sides gives \( \ln y = \ln(x^{x/2}) \). Applying the power rule of logarithms, this becomes \( \ln y = \frac{x}{2} \ln x \).

Step 2: Differentiate implicitly with respect to x

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

Step 3: Solve for the derivative

Re-arrange the derivative expression to solve for \( \frac{dy}{dx} \): \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} (\ln x + 1) \). Multiply both sides by \( y \) to isolate the derivative: \( \frac{dy}{dx} = y \left( \frac{1}{2}(\ln x + 1) \right) \). Substitute back \( y = x^{x/2} \) and the expression becomes \( \frac{dy}{dx} = x^{x/2} \cdot \frac{1}{2}(\ln x + 1) \).

Step 4: Simplify the expression

The derivative can be simplified to: \( \frac{dy}{dx} = \frac{x^{x/2}}{2} (\ln x + 1) \). This cleaned-up form clearly presents the derivative of the function \( y = \sqrt{x}^x \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus used for finding the derivative of functions that are not easily solved for one variable in terms of another. When dealing with expressions where both variables are interdependent, explicit separation into "y = f(x)" form may not be straightforward or even possible.

• The key idea is differentiating both sides of the equation simultaneously with respect to x, acknowledging that y implicitly depends on x.
• This method is beneficial when the equation involves operations like products or powers of the variable of interest.
• After differentiating, you aim to solve for \( \frac{dy}{dx} \) by treating derivatives as another term to solve for.

In the exercise, this approach allows us to handle both the complicated exponent and logarithmic forms efficiently.

Product Rule

The product rule is essential when differentiating expressions where two functions are multiplied together. If you have two functions, say \( u(x) \) and \( v(x) \), their product's derivative is calculated as follows:

\[ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

• Think of this as taking the derivative of the first function times the second, plus the first function times the derivative of the second.
• This rule keeps everything neatly organized, so you don't miss out on any part of the function terms.

In our exercise, the product \( \frac{x}{2} \cdot \ln x \) is differentiated using the product rule, resulting in a combination of terms from both components of the function.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a logarithm to the base e, where e is approximately 2.71828. It is extensively used in calculus due to its natural properties that simplify differentiation, particularly in logarithmic differentiation.

• When differentiating \( \ln x \), the result is simply \( \frac{1}{x} \).
• This behavior is attributed to the inverse relationship it shares with the exponential function \( e^x \).
• In expressions like \( \ln(y) = \ln(x^{x/2}) \), properties of logarithms simplify manipulation, such as using the power rule \( \ln(a^b) = b \cdot \ln a \).

Understanding and applying these properties streamlines the differentiation process, as seen in our step-by-step calculation.

Chain Rule

The chain rule in calculus helps when differentiating composite functions, those functions that could be expressed as one function inside another. It provides a structured way to take derivatives when operations are nested.
The formula is:

\[ \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \]

• This rule states that you differentiate the outer function, leaving the inside unchanged, and then multiply it by the derivative of the inside function.
• In our example, when differentiating \( \ln y \), the chain rule plays a key role in recognizing that y is actually a function of x. As such, we derive the expression respectably, resulting in \( \frac{1}{y} \cdot \frac{dy}{dx} \).

Without the chain rule, tackling such compositions would be exceedingly complex, especially in logarithmic contexts like the one presented in the exercise.