Problem 12 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ h(x)=\ln \left(x+\sqrt{x^{2}-1}\right) $$

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{\sqrt{x^2 - 1}}\).

Step 1: Identify the Outer Function

The function given is composed of an outer function and an inner function. The outer function is the natural logarithm, denoted as \( ext{ln}(...)\).

Step 2: Identify the Inner Function

The expression inside the logarithm, \(x + \sqrt{x^2 - 1}\), is the inner function. We need to differentiate this later.

Step 3: Differentiate the Outer Function

Using the chain rule, the derivative of \(h(x) = \ln(u)\), where \(u = x + \sqrt{x^2 - 1}\), is \(\frac{1}{u} \times \frac{du}{dx}\).

Step 4: Differentiate the Inner Function

Now differentiate \(u = x + \sqrt{x^2 - 1}\). The derivative of \(x\) is 1. Apply the derivative of a square root \(\sqrt{x^2 - 1}\), which is \(\frac{1}{2\sqrt{x^2 - 1}}\) times the derivative of \(x^2 - 1\).

Step 5: Apply the Chain Rule to the Inner Derivative

The derivative of \(x^2 - 1\) is \(2x\). Therefore, the derivative of \(\sqrt{x^2 - 1}\) is \(\frac{1 \cdot 2x}{2\sqrt{x^2 - 1}} = \frac{x}{\sqrt{x^2 - 1}}\).

Step 6: Combine the Derivatives

The derivative of \(u = x + \sqrt{x^2 - 1}\) is \(1 + \frac{x}{\sqrt{x^2 - 1}}\).

Step 7: Substitute Back into the Chain Rule

Substitute \(du/dx = 1 + \frac{x}{\sqrt{x^2 - 1}}\) into the chain rule expression: \(\frac{1}{x + \sqrt{x^2 - 1}} \cdot (1 + \frac{x}{\sqrt{x^2 - 1}})\).

Step 8: Simplify the Expression

The derivative simplifies to: \(\frac{1 + \frac{x}{\sqrt{x^2 - 1}}}{x + \sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1} + x}{(x^2 - 1)^{1/2}(x + \sqrt{x^2 - 1})}\). This simplifies further to \(\frac{1}{\sqrt{x^2 - 1}}\) as the final derivative.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiating composite functions. Imagine you have a function that is made up of other functions nested inside it. The chain rule helps you unravel these layers and find the derivative. This is done by differentiating each layer separately and then putting them all back together in a certain way.
To apply the chain rule, you first identify the outer and inner functions. For example, if you have a function like \( h(x) = \ln(g(x)) \), the logarithm \( \ln \) is the outer function, and \( g(x) \) is the inner function.

• Differentiate the outer function as if the inner function were just "x".
• Then, multiply that derivative by the derivative of the inner function.

This is expressed mathematically as: if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This approach simplifies complex differentiation problems by breaking them into smaller, manageable parts.

Natural Logarithm Differentiation

Differentiating natural logarithms is a common task in calculus. The natural logarithm function, denoted by \( \ln(x) \), is special because its rate of change is inversely proportional to \( x \).
The derivative of \( \ln(x) \) is straightforward: \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \). When you have a composite function inside the logarithm, like \( \ln(u(x)) \), you use the chain rule to differentiate it. The result is \( \frac{1}{u} \cdot \frac{du}{dx} \), which means you first take \( 1 \) over the entire inner function and then multiply by the derivative of the inner function.
This makes natural logarithms easier to handle. They just require careful differentiation of the function inside the logarithm and applying the chain rule effectively.

Inner and Outer Functions

In calculus, composite functions are common and are characterized by having inner and outer layers. Distinguishing between these layers is an essential skill when finding derivatives.
An outer function is the main operation or transformation applied to everything inside. For example, in \( \ln(x + \sqrt{x^2 - 1}) \), the natural logarithm \( \ln \) is the outer function.
The inner function is what you have within the outer function: \( x + \sqrt{x^2 - 1} \) in our example. Essentially, it's the argument of the outer function.

• To identify them, look for the larger structure of the function for the outer part.
• Everything inside that structure becomes the inner part.

Recognizing these layers simplifies applying the chain rule because you handle each part separately, making differentiation much more manageable.