Problem 53 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$ y=\sqrt{\frac{x-1}{x^{4}+1}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \sqrt{\frac{x-1}{x^{4}+1}} \cdot \frac{1}{2} \left( \frac{1}{x-1} - \frac{4x^{3}}{x^{4}+1} \right) \).

Step 1: Understand the Problem

We need to find the derivative of the function \( y = \sqrt{\frac{x-1}{x^{4}+1}} \). This can be done by using logarithmic differentiation.

Step 2: Express the Function with Exponents

Rewrite the function in an exponential form: \( y = \left(\frac{x-1}{x^{4}+1}\right)^{1/2} \). This will help us take the logarithm more easily.

Step 3: Take the Natural Logarithm

Take the natural logarithm of both sides: \( \ln y = \frac{1}{2} \ln \left(\frac{x-1}{x^{4}+1}\right) \). Use logarithmic properties to further simplify: \( \ln y = \frac{1}{2} ( \ln(x-1) - \ln(x^{4}+1) ) \).

Step 4: Differentiate Both Sides

Differentiate both sides with respect to \( x \). The left side becomes \( \frac{1}{y} \frac{dy}{dx} \), and the right side becomes \( \frac{1}{2} \left( \frac{1}{x-1} - \frac{4x^{3}}{x^{4}+1} \right) \).

Step 5: Solve for \( \frac{dy}{dx} \)

Solve the equation from the differentiation step for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = y \cdot \frac{1}{2} \left( \frac{1}{x-1} - \frac{4x^{3}}{x^{4}+1} \right) \).

Step 6: Substitute \( y \) Back

Substitute \( y = \sqrt{\frac{x-1}{x^{4}+1}} \) back into the expression for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \sqrt{\frac{x-1}{x^{4}+1}} \cdot \frac{1}{2} \left( \frac{1}{x-1} - \frac{4x^{3}}{x^{4}+1} \right) \).

Key Concepts

Derivative of Functions

To begin understanding derivatives, think of them as a tool that helps us find how much a function is changing at any given point. The derivative of a function gives us the slope of the tangent line at a specific point on the curve of the function. This is a fundamental concept in calculus, often used to find rates of change in real-life scenarios.
Here are some key points about derivatives:

• The derivative is denoted as \( \frac{dy}{dx} \) or \( f'(x) \) if the function is \( y = f(x) \).
• A positive derivative means the function is increasing, while a negative derivative indicates it's decreasing.
• If the derivative is zero, the function could be at a local maximum or minimum.

The process of finding a derivative can vary based on the form of the function, such as using the chain rule for composite functions or simplifying the function first by writing it in a different form, as we did by using an exponential form.

Exponential Form

The exponential form is a powerful tool when tackling calculus problems that involve complicated expressions. It often allows you to rewrite functions in a simpler form that is easier to manipulate while finding derivatives or integrals. For the given function, originally written under a square root, we transformed it as follows:
\[ y = \sqrt{\frac{x-1}{x^{4}+1}} = \left(\frac{x-1}{x^{4}+1}\right)^{1/2} \]
This expression allows us to apply logarithms, thereby simplifying the differentiation process. With the function now in exponential form, applying other mathematical operations, like taking logarithms, becomes more straightforward.
Remember, using the exponential form can make your work significantly clearer, revealing underlying patterns or allowing for effective application of rules and techniques like logarithmic differentiation.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), plays a crucial role in calculus. It's the inverse function of the exponential function \( e^x \). Using the natural logarithm can greatly simplify the process of finding derivatives of complex functions through logarithmic differentiation.
In our solution, we took the natural logarithm of both sides of the equation for ease of differentiation:
\[ \ln y = \frac{1}{2} ( \ln(x-1) - \ln(x^{4}+1) ) \]
Here, the properties of logarithms allow us to break apart the function into a difference of simpler logarithmic functions. This transformation simplifies differentiation because logarithmic derivatives are straightforward to compute:

• The derivative of \( \ln(u) \) is \( \frac{1}{u} \frac{du}{dx} \).

Recognize the power of natural logarithms in transforming complicated functions into manageable pieces for differentiation.

Chain Rule

The chain rule is essential when differentiating composite functions. A composite function is one where a function is applied to another function, like \( f(g(x)) \). The chain rule helps us find the derivative of these composite functions efficiently.
The basic formula of the chain rule is:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
This indicates that to differentiate a composite function, you multiply the derivative of the outer function by the derivative of the inner function. In our problem, we implicitly used the chain rule. When differentiating the natural logarithm expression, we implicitly differentiated the inner functions \( \frac{x-1}{x^4+1} \), ensuring correct differentiation across the entire composition.
Understanding and applying the chain rule allows you to handle even the most intricate functions by breaking them down into simpler parts.