Problem 58 - Thomas Calculus

Question

In Exercises \(57-70\) , use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt{\left(x^{2}+1\right)(x-1)^{2}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \sqrt{(x^2 + 1)(x - 1)^2} \left( \frac{x}{x^2 + 1} + \frac{1}{x - 1} \right) \).

Step 1: Take the Natural Logarithm

To simplify differentiation, start by taking the natural logarithm of both sides of the equation. Set \( y = \sqrt{(x^2 + 1)(x - 1)^2} \), then apply the logarithm: \( \ln(y) = \ln(\sqrt{(x^2 + 1)(x - 1)^2}) \).

Step 2: Simplify Using Logarithm Properties

Use the property \( \ln(a^b) = b \cdot \ln(a) \) and \( \ln(\sqrt{a}) = \frac{1}{2}\ln(a) \) to simplify: \( \ln(y) = \frac{1}{2} \ln((x^2 + 1)(x - 1)^2) \). This further becomes \( \ln(y) = \frac{1}{2} \left( \ln(x^2 + 1) + 2 \ln(x - 1) \right) \).

Step 3: Differentiate Both Sides

Differentiate both sides with respect to \( x \). The left side gives \( \frac{d}{dx}[\ln(y)] = \frac{1}{y} \frac{dy}{dx} \). For the right side, differentiate: \( \frac{1}{2} \left( \frac{2x}{x^2 + 1} + \frac{2}{x - 1} \right) \).

Step 4: Solve for \( \frac{dy}{dx} \)

Combine terms: \( \frac{1}{y} \frac{dy}{dx} = \frac{x}{x^2 + 1} + \frac{1}{x - 1} \). Multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \): \( \frac{dy}{dx} = y \left( \frac{x}{x^2 + 1} + \frac{1}{x - 1} \right) \).

Step 5: Substitute \( y \) Back

Recall that \( y = \sqrt{(x^2 + 1)(x - 1)^2} \), substitute back to find \( \frac{dy}{dx} = \sqrt{(x^2 + 1)(x - 1)^2} \left( \frac{x}{x^2 + 1} + \frac{1}{x - 1} \right) \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus that describe how a function changes as its input changes. Imagine driving a car; the speedometer measures the car's speed, or how quickly its position changes over time. In math, derivatives measure similar kinds of changes.

Here's the gist:

• A derivative calculates the slope of a curve at any given point, giving an instantaneous rate of change.
• In equations, it's represented as \( f'(x) \) or \( \frac{dy}{dx} \), where \( y \) is a function of \( x \).
• To find the derivative, we apply specific rules and techniques, like the power rule, product rule, or chain rule, depending on the function's form.

In the context of logarithmic differentiation, derivatives help us find rates of change for functions that are complex, involving products, quotients, or even roots. Here, they are simplified using logarithmic identities before differentiation is applied.

Exploring the Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \), is a special type of logarithm with the base \( e \), where \( e \approx 2.71828 \), a very important number in mathematics due to its unique properties.

Here's why it matters:

• The natural logarithm helps to transform multiplicative relationships into additive ones, which are much easier to work with mathematically.
• Common properties include \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(a^b) = b\ln(a) \), making it useful for simplifying complicated expressions before differentiating.
• When dealing with logarithmic differentiation, we use \( \ln \) because it transforms the original function into a form that's easier to differentiate, employing these properties.

In our specific problem, the function \( y = \sqrt{(x^2 + 1)(x - 1)^2} \) turns into a more manageable expression by applying \( \ln \) to both sides.

Applying Differentiation Rules

Differentiation rules are the tools we'll use to find the derivatives. Each type of function or expression has its corresponding rule, which simplifies the process of finding the derivative of a given function.

Let's go over some essential rules:

• The **power rule** is used for functions of the form \( x^n \) and states \( \frac{d}{dx}[x^n] = nx^{n-1} \).
• The **product rule** is applied to functions multiplied together: \( \frac{d}{dx}[uv] = u'v + uv' \).
• The **chain rule** is for composite functions, allowing us to differentiate the outer function and then multiply by the derivative of the inner: \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \).

In the exercise, after employing logarithmic properties and simplifying, we differentiated using these rules step-by-step, showing how they combine effectively. This approach gives us the toolset to handle even more complex expressions by "breaking them down" into simpler forms and applying the rules more easily.