Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ h^{\prime}(x) \text { if } h(x)=\ln \left(x+\sqrt{x^{2}-1}\right) $$

Step-by-Step Solution

Answer

\( h'(x) = \frac{1}{\sqrt{x^2 - 1}} \)

Step 1: Identify the problem type

The problem is to find the derivative of a logarithmic function composed with a square root function. This requires the use of the chain rule and the derivative of the natural logarithm.

Step 2: Rewrite using substitution

Let's simplify the expression inside the logarithm by setting it as a new function: let \( u(x) = x + \sqrt{x^2 - 1} \). Now we have \( h(x) = \ln(u(x)) \).

Step 3: Differentiate \( h(x) \) using the chain rule

According to the chain rule, the derivative of \( h(x) \) is \( h'(x) = \frac{1}{u(x)} \cdot u'(x) \), because the derivative of \( \ln(u) \) is \( \frac{1}{u} \).

Step 4: Find the derivative of \( u(x) \)

Differentiate \( u(x) = x + \sqrt{x^2 - 1} \). For the derivative of \( \sqrt{x^2 - 1} \), use the chain rule: \( \frac{d}{dx}\sqrt{x^2 - 1} = \frac{1}{2\sqrt{x^2 - 1}} \times 2x = \frac{x}{\sqrt{x^2 - 1}} \). Thus, \( u'(x) = 1 + \frac{x}{\sqrt{x^2 - 1}} \).

Step 5: Substitute \( u'(x) \) back into the derivative of \( h(x) \)

Substitute \( u'(x) = 1 + \frac{x}{\sqrt{x^2 - 1}} \) into \( h'(x) = \frac{1}{u(x)} \cdot u'(x) \). This gives \( h'(x) = \frac{1}{x + \sqrt{x^2 - 1}} \times \left(1 + \frac{x}{\sqrt{x^2 - 1}}\right) \).

Step 6: Simplify the expression

Simplify \( h'(x) = \frac{1}{x + \sqrt{x^2 - 1}} \times \left(\frac{\sqrt{x^2 - 1} + x}{\sqrt{x^2 - 1}}\right) \). This simplifies further to \( h'(x) = \frac{\sqrt{x^2 - 1} + x}{(x + \sqrt{x^2 - 1})\sqrt{x^2 - 1}} \). Since \( (\sqrt{x^2 - 1} + x) = \frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} \), the expression simplifies to \( \frac{1}{\sqrt{x^2 - 1}} \).

Step 7: Present Final Answer

The derivative of the given function is \( h'(x) = \frac{1}{\sqrt{x^2 - 1}} \).

Key Concepts

Chain Rule Calculus

In calculus, the chain rule is an essential technique for differentiating composite functions. Here's a simple breakdown: When you have a function inside another function, like a function of another function's form, you need to use the chain rule. Think of it like peeling an onion, layer by layer.

In mathematical notation, if you have a function \( y = f(g(x)) \), the derivative \( y' \) is determined by multiplying the derivative of the outer function by the derivative of the inner function:

• Derive the outer function \( f(g(x)) \) with respect to \( g(x) \).
• Multiply it by the derivative of the inner function \( g(x) \) with respect to \( x \).

This approach applies perfectly in our problem, where the natural logarithm \( \ln(u(x)) \) is the outer function and the expression inside the logarithm is \( u(x) = x + \sqrt{x^2 - 1} \) as the inner function.

Understanding and practicing the chain rule can significantly simplify tackling composite functions in calculus problems, vital for deriving more complex functions. By leveraging this rule, we can break down complicated derivatives into manageable parts, allowing for precise and efficient solutions.

Natural Logarithm Derivatives

The natural logarithm function, denoted as \( \ln(x) \), is a fundamental concept in calculus with specific differentiation properties. The derivative of the natural logarithm function is unique and simple:
The derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This means whenever you derive \( \ln(x) \), you essentially 'invert' \( x \).

However, if the argument of the logarithm is a more complex function \( u(x) \), you would first consider the derivative of \( \ln(u(x)) \), which follows the chain rule. The derivative is:\[ \frac{1}{u(x)} \times u'(x) \]
In our exercise, utilizing the property of natural logarithms allows us to differentiate \( \ln(x + \sqrt{x^2 - 1}) \) by using this formula. By evaluating the inner function \( u(x) = x + \sqrt{x^2 - 1} \), and subsequently its derivative, we simplify the derivative calculation.

This property is particularly useful when dealing with logarithmic differentiation tasks, providing clarity and consistency in solving complex derivatives.

Differentiation Techniques

Differentiation techniques are the cornerstone of calculus, allowing us to find how a function's output changes with its input. Each technique suits different forms and relationships among functions, notably when handling composite or complex functions.

Some techniques include:

• Power Rule: For \( f(x) = x^n \), the derivative \( f'(x) = nx^{n-1} \).
• Product Rule: When differentiating products, \((uv)' = u'v + uv' \).
• Quotient Rule: Handles quotients like \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

In particular, the chain rule and natural logarithm differentiation are pivotal here.
When working with \( h(x) = \ln(x + \sqrt{x^2 - 1}) \), these combined techniques help dissect the problem. The inner function \( x + \sqrt{x^2 - 1} \) involves the power rule within its square root component, followed by careful application of chain rule calculus.

Understanding and applying differential rules proficiently tailors our approach to tackle each function's unique problem, ensuring accurate results in a structured manner. Mastery over these differentiation tricks equips one to handle even the most intricately nested functions with ease.