Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ f^{\prime}(1) \text { if } f(x)=x^{\sin x} $$

Step-by-Step Solution

Answer

The derivative \( f'(1) = \sin 1 \).

Step 1: Express the function in logarithmic form

Given the function \( f(x) = x^{\sin x} \), we start by taking the natural logarithm of both sides to simplify the expression for differentiation: \( \ln f(x) = \ln(x^{\sin x}) \). Applying the logarithmic identity, this becomes \( \ln f(x) = \sin x \cdot \ln x \).

Step 2: Differentiate using implicit differentiation

Differentiate both sides with respect to \( x \). The left side becomes \( \frac{1}{f(x)} f'(x) \). The right side requires the product rule: \( \frac{d}{dx} (\sin x \cdot \ln x) = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \). Thus, we have: \[ \frac{1}{f(x)} f'(x) = \cos x \cdot \ln x + \frac{\sin x}{x} \].

Step 3: Solve for \( f'(x) \)

Multiply both sides by \( f(x) = x^{\sin x} \) to isolate \( f'(x) \): \[ f'(x) = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \].

Step 4: Calculate \( f'(1) \)

Substitute \( x = 1 \) into the expression for \( f'(x) \): \[ f'(1) = 1^{\sin 1} \left( \cos 1 \cdot \ln 1 + \frac{\sin 1}{1} \right) \]. Since \( \ln 1 = 0 \), the expression simplifies to: \[ f'(1) = 1 \cdot \left( 0 + \sin 1 \right) = \sin 1 \].

Key Concepts

Implicit Differentiation

When differentiating functions, sometimes we encounter equations that don't allow us to solve for one variable directly in terms of another, or it's too complex. Implicit differentiation offers a clever way around this.Instead of solving for one variable, we differentiate both sides of an equation with respect to the independent variable. This often involves using the chain rule. In this problem, after expressing the function using a logarithm, we applied implicit differentiation. We took the derivative on both sides of the equation \( \ln f(x) = \sin x \cdot \ln x \) with respect to \( x \). This allowed us to eventually find \( f'(x) \), even though \( f(x) \) wasn't solved as a simple expression.

Product Rule

The product rule is a vital tool in calculus when dealing with the derivative of a product of two functions. It states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is:\[ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]In our function, \( \ln f(x) = \sin x \cdot \ln x \), we recognized that the right side needed the product rule for differentiation. We let \( u(x) = \sin x \) and \( v(x) = \ln x \). Using the product rule, the derivative \( \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \) was found.

Logarithmic Differentiation

Logarithmic differentiation takes advantage of logarithms to simplify differentiation, especially when tackling complex functions like \( x^{\sin x} \).By taking the natural logarithm of such functions, we can transform products into sums or powers into multipliers, which are easier to differentiate. In this case, transforming the original function to \( \ln f(x) = \sin x \cdot \ln x \) allowed differentiation using simpler techniques, such as the product rule. After differentiation, it’s straightforward to solve for \( f'(x) \). Logarithmic differentiation is particularly handy for functions where the variable is in both the base and the exponent.

Exponential Functions

Exponential functions are a class of functions where a constant base is raised to a variable exponent, such as \( a^x \). However, in this exercise, the function \( x^{\sin x} \) is a slightly unusual form, known as a variable base with a variable exponent.These types of functions often require special techniques like logarithmic differentiation, as shown, to find their derivatives. Exponential functions have powerful growth and decay properties and appear frequently in real-world contexts, such as finance, biology, and physics. When working with them in calculus, thoroughly understanding differentiation techniques, like those used here, is crucial for analyzing and interpreting such functions effectively.