Problem 9 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ g(x)=\ln \frac{a-x}{a+x} $$

Step-by-Step Solution

Answer

The derivative of \( g(x) = \ln \frac{a-x}{a+x} \) is \( \frac{-2a}{(a-x)(a+x)} \).

Step 1: Identify the form of the function

The function given is of the form \( g(x) = \ln \left( \frac{a-x}{a+x} \right) \). This is a natural logarithm applied to a fractional expression.

Step 2: Apply the Chain Rule

To differentiate the given function, use the chain rule. The chain rule states that if \( u(x) \) is a function inside another function, such as \( \ln(u(x)) \), then \( \frac{d}{dx}\ln(u(x)) = \frac{1}{u(x)} \cdot u'(x) \). Here, \( u(x) = \frac{a-x}{a+x} \).

Step 3: Differentiate the Inner Function

Differentiate \( u(x) = \frac{a-x}{a+x} \) using the quotient rule. The quotient rule is \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \) where \( f(x) = a-x \) and \( g(x) = a+x \). Thus, \( f'(x) = -1 \) and \( g'(x) = 1 \).

Step 4: Apply the Quotient Rule

Using the quotient rule on \( u(x) \), we get:\[ u'(x) = \frac{(-1)(a+x) - (a-x)(1)}{(a+x)^2} = \frac{-(a+x) - (a-x)}{(a+x)^2} = \frac{-2a}{(a+x)^2}\].

Step 5: Differentiate the Outer Function

The derivative of \( g(x) = \ln(u(x)) \) is given by:\[ g'(x) = \frac{1}{\frac{a-x}{a+x}} \cdot \frac{-2a}{(a+x)^2}\]Simplifying, this becomes:\[ g'(x) = \frac{(a+x)}{a-x} \cdot \frac{-2a}{(a+x)^2} = \frac{-2a}{(a-x)(a+x)}\].

Step 6: Simplify the Derivative

Upon simplifying further, the expression \( \frac{-2a}{(a-x)(a+x)} \) is already in a simplified form, and corresponds to the final answer of the derivative of \( g(x) \).

Key Concepts

Chain Rule

The chain rule is essential when dealing with composite functions, where one function is nested inside another. In simpler terms, if you have a function like \( h(x) = f(g(x)) \), the chain rule helps compute the derivative of this composition. The rule states that the derivative \( h'(x) \) is calculated as the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
For instance, if you're dealing with \( g(x) = \ln(u(x)) \), you can treat \( \ln \) as the outer function and \( u(x) \) as the inner function. Hence, by applying the chain rule:

• Derivative of the outer function: \( \frac{d}{du}\ln(u) = \frac{1}{u} \)
• Derivative of the inner function \( u(x) \)

By using this knowledge, you substitute \( u'(x) \), the derivative of \( u(x) \), back into the equation to find the full derivative of the composite \( g(x) \). The chain rule elegantly allows handling these powerful transformations.

Quotient Rule

The quotient rule is invaluable for differentiating functions that are in the form of one function divided by another, say \( u(x) = \frac{f(x)}{g(x)} \). When you're tasked with finding the derivative of such a ratio, the quotient rule is the go-to tool.
It states:

• The derivative \( (\frac{f}{g})' = \frac{f'g - fg'}{g^2} \)

Applying this in practice can be simplified by remembering how you need to take the derivative of the numerator (\( f'(x) \)) and the derivative of the denominator (\( g'(x) \)) separately, then apply them within the formula.
For example, with \( u(x) = \frac{a-x}{a+x} \), the steps involve:

• Finding \( f'(x) = -1 \) for \( f(x) = a - x \)
• Finding \( g'(x) = 1 \) for \( g(x) = a + x \)
• Substituting these into the quotient rule formula

Ultimately, this leads to the derivative \( u'(x) \), which is critical when further used with the chain rule as seen in differentiating \( g(x) \).

Natural Logarithm

The natural logarithm, often written as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \) is approximately 2.71828. It's a fundamental concept in calculus due to its unique properties, especially when differentiating.
When differentiating \( \ln(x) \), you can simply derive it as \( \frac{1}{x} \). This property becomes very useful when logarithmic functions are part of more complex expressions, like those wrapped around fractions or other functions.

• Key property: \( \frac{d}{dx}\ln(x) = \frac{1}{x} \)
• Applied to composite functions such as \( \ln(u(x)) \)

The combination of logarithmic differentiation with other rules, such as the chain and quotient rules, facilitates managing the complexity of rational and nested functions. For instance, in our exercise, identifying \( \ln \) as the outer layer of the \( g(x) \) function simplifies the differentiation, breaking it down into a more digestible format.