Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find and simplify \(f^{\prime}(1)\) if $$ f(x)=\ln \left(\frac{a x-b}{a x+b}\right)^{c}, \text { where } c=\frac{a^{2}-b^{2}}{2 a b} . $$

Step-by-Step Solution

Answer

The simplified value of \( f'(1) \) is 1.

Step 1: Differentiate the Function

Given the function \( f(x) = \ln \left(\frac{a x-b}{a x+b}\right)^{c} \), use the chain rule and the property of logarithms, \( \ln(u^c) = c \cdot \ln(u) \). Set \( u = \frac{a x - b}{a x + b} \). The derivative is \( f'(x) = c \cdot \frac{d}{dx} \ln(u) = c \cdot \frac{1}{u} \cdot \frac{du}{dx} \).

Step 2: Differentiate the Inner Function u

Find \( \frac{du}{dx} \) where \( u = \frac{a x - b}{a x + b} \). Use the quotient rule: \( \frac{d}{dx} \left( \frac{v}{w} \right) = \frac{v'w - vw'}{w^2} \), where \( v = a x - b \) and \( w = a x + b \). Compute \( v' = a \) and \( w' = a \), so \( \frac{du}{dx} = \frac{(a)(a x + b) - (a x - b)(a)}{(a x + b)^2} = \frac{2ab}{(a x + b)^2} \).

Step 3: Substitute Back into the Derivative

Now substitute \( \frac{du}{dx} \) back into the expression for \( f'(x) \). This gives \( f'(x) = c \cdot \frac{1}{\frac{a x - b}{a x + b}} \cdot \frac{2ab}{(a x + b)^2} = c \cdot \frac{2ab(a x + b)}{(a x - b)(a x + b)^2} \). Simplifying gives \( f'(x) = \frac{2abc}{(a x - b)(a x + b)} \).

Step 4: Substitute c and Evaluate at x = 1

We have \( c = \frac{a^2 - b^2}{2ab} \). Substitute \( c \) into \( f'(x) \): \( f'(x) = \frac{2ab \cdot \frac{a^2-b^2}{2ab}}{(a x-b)(a x+b)} = \frac{a^2-b^2}{(a x-b)(a x+b)} \). Evaluate \( f'(1) \): \( f'(1) = \frac{a^2-b^2}{(a-b)(a+b)} \).

Step 5: Simplify the Expression

Notice that \( f'(1) = \frac{a^2 - b^2}{a^2 - b^2} = 1 \). Thus, the simplified value of \( f'(1) \) is 1.

Key Concepts

Understanding the Chain Rule

When we deal with derivatives, the chain rule is essential for differentiating composite functions. Composite functions are functions where one function is inside of another function, much like nested layers. In the given problem, the logarithm function \[ f(x) = \ln \left( \frac{a x-b}{a x+b} \right)^{c} \]is such a composite function. The chain rule helps in breaking down the differentiation of such layers. The rule states:

• If you have a function \( g(h(x)) \), its derivative is \( g'(h(x)) \cdot h'(x) \).

In the exercise, we replace the inner part with a simpler expression, \( u = \frac{a x - b}{a x + b} \), making it easier to differentiate. With the chain rule, once \( u \) is assigned, we find the derivative of the log function in terms of \( u \) first, then multiply by the derivative of \( u \) itself. This layered approach simplifies the work of differentiation, letting us handle complex expressions one step at a time.

Applying the Quotient Rule

The quotient rule is a key technique used when differentiating functions of the form \( \frac{v(x)}{w(x)} \). It allows us to find the derivative when functions are divided by each other. Recall from the original exercise, the function \( u = \frac{a x - b}{a x + b} \) appears. To differentiate \( u \), we apply the quotient rule:

• \( \frac{d}{dx} \left( \frac{v}{w} \right) = \frac{v'w - vw'}{w^2} \)

Here, we have:

• \( v = a x - b \)
• \( w = a x + b \)
• \( v' = a \)
• \( w' = a \)

This results in:

• \( \frac{du}{dx} = \frac{(a)(a x + b) - (a x - b)(a)}{(a x + b)^2} \)
• Simplifying yields \( \frac{du}{dx} = \frac{2ab}{(a x + b)^2} \)

Using the quotient rule helps manage the algebra involved when differentiating by keeping the expressions neat and systematic.

Exploring Logarithmic Differentiation

Logarithmic differentiation is a technique that simplifies the process of finding derivatives, especially when exponents are involved. In our example, the expression is initially given as \( \ln \left( \frac{a x-b}{a x+b} \right)^{c} \).We use the property of logarithms that allows us to bring down the exponent for easier differentiation:

• \( \ln(u^c) = c \cdot \ln(u) \).

This property turns our initial complicated logarithmic expression into something more manageable:

• \( f(x) = c \cdot \ln \left( \frac{a x-b}{a x+b} \right) \).

It converts the power into a factor, making differentiation simpler. Once it's expressed in this form, you can apply standard differentiation rules, like the chain rule, to find the derivative. This method is particularly powerful when the original function has complex exponents, streamlining calculations and reducing errors.

Simplification in Calculus

Simplification is a crucial final step in solving calculus problems, ensuring we present results in their simplest and most understandable form. In the original exercise, after deriving the function using the chain rule, quotient rule, and logarithmic differentiation, we had:\[ f'(x) = \frac{2abc}{(a x - b)(a x + b)}. \]By substituting \( c = \frac{a^2 - b^2}{2ab} \), the expression reduced to:\[ f'(x) = \frac{a^2-b^2}{(a x-b)(a x+b)}. \]Further evaluation at \( x = 1 \) yields:\[ f'(1) = \frac{a^2 - b^2}{(a-b)(a+b)} = 1, \]which is the simplest form. Simplification involves canceling common factors and reducing complex fractions, making the expression easy to interpret and often revealing meaningful insights. Think of it as the polishing step, where we ensure the math we present is clear and precise.