Problem 30 - Brief Applied Calculus

Question

Find the second derivative of each function. $$ \frac{x}{x^{2}-1} $$

Step-by-Step Solution

Answer

The second derivative of \( \frac{x}{x^2 - 1} \) is \( \frac{-6x^5 + 4x^3 + 2x}{(x^2 - 1)^4} \)."

Step 1: Identify the function and differentiate

The given function is \( f(x) = \frac{x}{x^2 - 1} \). We need to find its first derivative. The function \( f(x) \) is a quotient, so we'll apply the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = x \) and \( v = x^2 - 1 \).

Step 2: Apply the Quotient Rule

Differentiate \( u \) and \( v \): \( u = x \) implies \( u' = 1 \). For \( v = x^2 - 1 \), \( v' = 2x \). Applying the quotient rule: \[\frac{d}{dx} \left( \frac{x}{x^2 - 1} \right) = \frac{1 \cdot (x^2 - 1) - x \cdot 2x}{(x^2 - 1)^2} = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2}.\]

Step 3: Simplify the first derivative

The expression simplifies to: \( f'(x) = \frac{-x^2 - 1}{(x^2 - 1)^2} \). For clarity and easier differentiation, rewrite it as \( f'(x) = \frac{-1(x^2 + 1)}{(x^2 - 1)^2} \).

Step 4: Differentiate to find the second derivative

We need to differentiate \( f'(x) = \frac{-1(x^2 + 1)}{(x^2 - 1)^2} \) again. Apply the quotient rule with \( u = -1(x^2 + 1) \) and \( v = (x^2 - 1)^2 \). Find \( u' = -2x \) and \( v' = 2(x^2 - 1)(2x) = 4x(x^2 - 1) \).

Step 5: Apply the Quotient Rule for second derivative

Use the quotient rule: \[f''(x) = \frac{(-2x)(x^2-1)^2 - (-1)(x^2+1)(4x(x^2-1))}{(x^2-1)^4}.\] Simplify each part separately:- \((-2x)(x^2-1)^2 = -2x(x^4 - 2x^2 + 1) = -2x^5 + 4x^3 - 2x\)- \(4x(x^2+1)(x^2-1) = 4x(x^4 - x^2 + x^2 - 1) = 4x(x^4 - 1) = 4x^5 - 4x\)\(f''(x) = \frac{-2x^5 + 4x^3 - 2x - 4x^5 + 4x}{(x^2 - 1)^4} = \frac{-6x^5 + 4x^3 + 2x}{(x^2 - 1)^4}.\)

Step 6: Final simplified form of the second derivative

Simplify the expression: \[f''(x) = \frac{-6x^5 + 4x^3 + 2x}{(x^2 - 1)^4}\]This is the second derivative of the given function.

Key Concepts

Quotient Rule

In calculus, the Quotient Rule is a method used to differentiate functions that are expressed as one function divided by another. It's essential when you're dealing with rational functions where one function is on top of another. If you have a function in the form \( \frac{u}{v} \), the derivative of this function, denoted \( \left( \frac{u}{v} \right)' \), can be found using the formula:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

This allows you to find how the slope of the graph of the function changes, based on its inputs. In our exercise, \( u = x \) and \( v = x^2 - 1 \), and using the Quotient Rule helps you identify the rate at which the entire fraction is changing, by considering both \( u \) and \( v \) separately. Remember to differentiate \( u \) and \( v \) separately, and then substitute back into the formula.

First Derivative

The First Derivative of a function gives us significant information about the function's behavior, such as where it is increasing or decreasing. By applying the Quotient Rule to \( f(x) = \frac{x}{x^2 - 1} \), we first differentiate the numerator \( u = x \) to get \( u' = 1 \). We then differentiate the denominator \( v = x^2 - 1 \) to get \( v' = 2x \).

• Plugging these into the Quotient Rule formula gives us: \[ \frac{1(x^2 - 1) - x(2x)}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2} \]

This result shows how the original function changes at every point \( x \). The first derivative is crucial for identifying critical points which can tell us where the function might have local minima or maxima.

Second Derivative

The Second Derivative of a function provides additional information about the curvature of the function's graph. In this context, curvature refers to whether a graph looks like a valley (concave up) or a hill (concave down). To find it, we take the derivative of our first derivative \( f'(x) \).

• Here, we have \( f'(x) = \frac{-1(x^2 + 1)}{(x^2 - 1)^2} \).
• Using the Quotient Rule again, but this time with \( u = -1(x^2 + 1) \) and \( v = (x^2 - 1)^2 \), we find \( f''(x) \).
• The process involves careful differentiation and algebraic simplifications to maintain accuracy. Once simplified, the result is \[ f''(x) = \frac{-6x^5 + 4x^3 + 2x}{(x^2 - 1)^4} \].

This second derivative informs us about the inflection points and the nature of the curve, indicating regions of concavity.

Simplification in Calculus

Simplification is a critical part of working with derivatives, especially when dealing with complex expressions. It involves rewriting expressions to make them easier to understand or work with. In our exercise, simplifying the expression after applying the Quotient Rule is essential to get the correct result without unnecessary complexity.

• By simplifying \( f'(x) \) to \( \frac{-1(x^2 + 1)}{(x^2 - 1)^2} \), we make subsequent calculations more manageable.
• Furthermore, simplifying terms within \( f''(x) \) ensures clarity and reduces the chance of errors when interpreting results.

Careful algebraic manipulation and combining like terms, such as terms of similar degrees in contributions of \( x \), are key steps. This attention to simplicity aids not just in computation but also enhances understanding of the relationships within the function.