Problem 18 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ H(z)=\ln \sqrt{\frac{a^{2}-z^{2}}{a^{2}+z^{2}}} $$

Step-by-Step Solution

Answer

The derivative is \( H'(z) = \frac{2z^3}{(a^2-z^2)(a^2+z^2)} \).

Step 1: Recognize the Function Form

The given function is a natural logarithm of a square root. Recall the identity for square roots, where \( \sqrt{x} \) can be rewritten as \( x^{1/2} \). In this case, apply the rule inside the logarithm: \( \ln \sqrt{x} = \frac{1}{2} \ln x \). Thus, we rewrite the function as:\[ H(z) = \frac{1}{2} \ln \left( \frac{a^2-z^2}{a^2+z^2} \right) \]

Step 2: Use the Quotient Property of Logarithms

Apply the properties of logarithms, specifically the quotient property, which states \( \ln \left( \frac{x}{y} \right) = \ln x - \ln y \). Thus, apply this to our expression:\[ H(z) = \frac{1}{2} (\ln (a^2-z^2) - \ln (a^2+z^2)) \]

Step 3: Differentiate the Function

To differentiate \( H(z) \), we need to apply the chain rule and the derivative of the natural logarithm. The derivative of \( \ln u \) is \( \frac{1}{u} \cdot u' \). Differentiate each term separately:- First term: \( \frac{d}{dz} \left( \ln(a^2-z^2) \right) = \frac{-2z}{a^2-z^2} \) - Second term: \( \frac{d}{dz} \left( \ln(a^2+z^2) \right) = \frac{2z}{a^2+z^2} \)Now use these in the expression for \( H'(z) \):\[ H'(z) = \frac{1}{2} \left( \frac{-2z}{a^2-z^2} - \frac{2z}{a^2+z^2} \right) \]

Step 4: Simplify the Expression

Combine both parts within the differentiation by factoring out common terms:\[ H'(z) = \frac{z}{a^2-z^2} - \frac{z}{a^2+z^2} \]Combine the fractions:\[ H'(z) = z \left( \frac{1}{a^2-z^2} - \frac{1}{a^2+z^2} \right) \]Combine into a single fraction:\[ H'(z) = \frac{z \left( (a^2+z^2) - (a^2-z^2) \right)}{(a^2-z^2)(a^2+z^2)} \]\[ H'(z) = \frac{z \cdot 2z^2}{(a^2-z^2)(a^2+z^2)} \]\[ H'(z) = \frac{2z^3}{(a^2-z^2)(a^2+z^2)} \]

Key Concepts

Natural Logarithm Differentiation

Natural logarithms, often written as \( \ln(x) \), have specific rules when it comes to differentiation. The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \cdot u' \), where \( u' \) is the derivative of \( u \) with respect to the variable of differentiation. This is a crucial part of solving calculus problems involving logs.

When differentiating \( \ln \) functions, it is necessary to understand what is nested inside the logarithm. This is because the chain rule will often come into play, requiring that you take the derivative of the inside function separately. For example, if \( u = a^2 - z^2 \), its derivative \( u' \) would be \( -2z \).

• Knowing these derivatives now allows us to use the natural logarithm differentiation rule meticulously.
• The process involves finding \( u' \) first, then applying it to the main differentiation formula.

By breaking down the function using logarithmic differentiation rules, you can handle complex expressions in a structured manner.

Chain Rule

The chain rule is pivotal when differentiating composite functions, such as when a logarithm contains an expression sensitive to changes in \( z \). The rule states that if a function \( y = f(g(z)) \) is composite, its derivative is \( y' = f'(g(z)) \cdot g'(z) \). Simply put, you differentiate the outer function and multiply by the derivative of the inner function.

This exercise demonstrates the chain rule applied within the context of a logarithmic function. Inside our natural log function are terms like \( a^2 - z^2 \) and \( a^2 + z^2 \). These require differentiation by:\[ \frac{d}{dz}(a^2 - z^2) = -2z \] and\[ \frac{d}{dz}(a^2 + z^2) = 2z \]

Once each term is differentiated separately, the results are used according to the chain rule in executing the necessary logarithmic differentiation.

• Break down complex functions into simpler parts.
• Apply the chain rule for each part to find derivatives accurately.

Understanding and applying the chain rule is imperative for success in differential calculus.

Quotient Property of Logarithms

When dealing with logarithms, the quotient property is a key player in simplifying expressions. This property asserts that the logarithm of a quotient is equal to the difference of the logarithms. Mathematically, \( \ln \left( \frac{x}{y} \right) = \ln x - \ln y \). This powerful tool is incredibly handy for making complex logarithmic differentiations more manageable.

In our problem, the original function within the logarithm is a quotient:\[ \ln \left( \frac{a^2 - z^2}{a^2 + z^2} \right) \]
By applying the quotient property, we break it down into two separate parts that are easier to handle individually during differentiation:

\[ \ln(a^2 - z^2) - \ln(a^2 + z^2) \]
This step simplifies the subsequent application of logarithmic differentiation rules, especially when combined with the chain rule.

• Break down complicated quotients using this property.
• Pair with the chain rule and natural log differentiation for full solutions.

Simplifying logarithmic expressions using the quotient property can significantly ease solving differential calculus problems.