Problem 29 - Brief Applied Calculus

Question

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

Step-by-Step Solution

Answer

The second derivative is \( f''(x) = \frac{-2x(3-x^2)}{(x^2+1)^3} \).

Step 1: Understand the Function and Set the Goal

We are given the function \( f(x) = \frac{x}{x^2+1} \) and our task is to find its second derivative, which means we first need to find the first derivative and then differentiate again to obtain the second derivative, \( f''(x) \).

Step 2: Differentiate the Function Using the Quotient Rule

To find the first derivative, \( f'(x) \), of the function \( f(x) = \frac{x}{x^2+1} \), we use the quotient rule. The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = x \) and \( v = x^2 + 1 \).1. Find \( u' = 1 \).2. Find \( v' = 2x \).3. Substitute into the quotient rule: \[ f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} \] \[ = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} \] \[ = \frac{1 - x^2}{(x^2 + 1)^2} \].

Step 3: Differentiate Again to Find the Second Derivative

Now we find the second derivative \( f''(x) \) by differentiating \( f'(x) = \frac{1-x^2}{(x^2+1)^2} \) again, using the quotient rule:1. Let \( u = 1 - x^2 \) and \( v = (x^2 + 1)^2 \).2. Find \( u' = -2x \). To find \( v' \), use the chain rule: \[ v = (x^2 + 1)^2 \rightarrow v' = 2(x^2 + 1) \cdot (2x) = 4x(x^2 + 1) \].3. Substitute into the quotient rule: \[ f''(x) = \frac{(-2x)(x^2+1)^2 - (1-x^2)(4x(x^2+1))}{((x^2+1)^2)^2} \] \[ = \frac{-2x(x^2+1)^2 - 4x(1-x^2)(x^2+1)}{(x^2+1)^4} \].4. Simplify the expression above to find \( f''(x) \).

Step 4: Simplify the Second Derivative Expression

Simplifying \( f''(x) = \frac{-2x(x^2+1)^2 - 4x(1-x^2)(x^2+1)}{(x^2+1)^4} \):1. Factor out \(-2x(x^2+1)\): \[ = \frac{-2x(x^2+1)((x^2+1) + 2(1-x^2))}{(x^2+1)^4} \] \[ = \frac{-2x(x^2+1)(x^2 + 1 + 2 - 2x^2)}{(x^2+1)^4} \] \[ = \frac{-2x(x^2+1)(3-x^2)}{(x^2+1)^4} \].2. Simplifying further gives: \[ f''(x) = \frac{-2x(3-x^2)}{(x^2+1)^3} \].

Step 5: Conclusion: Write the Final Expression for the Second Derivative

The second derivative of the function \( f(x) = \frac{x}{x^2+1} \) is \( f''(x) = \frac{-2x(3-x^2)}{(x^2+1)^3} \).

Key Concepts

Quotient Rule

The quotient rule in calculus is pivotal when you're trying to differentiate a function that is presented as a fraction, specifically the ratio of two functions. It's a tool designed to work with expressions formatted as \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \). The formula for the quotient rule is:

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

In practice, this means you take the derivative of the numerator function \( u \), denoted as \( u' \), and multiply it by the original denominator \( v \). Then, subtract the product of the original numerator \( u \) and the derivative of the denominator \( v' \). Divide this difference by the square of the original denominator \( v^2 \).

This rule comes in very handy and is a must-know when tackling problems like finding the derivative of \( \frac{x}{x^2+1} \), where both the numerator and the denominator need to be handled separately and with precision.

Chain Rule

The chain rule is another critical technique in differentiation, often used when differentiating composite functions. It offers a way to manage functions nested within each other, like \( (x^2+1)^2 \).

Following a simple logic, if you have a function \( y = g(f(x)) \), the chain rule states that the derivative \( y' \) with respect to \( x \) is:

• \( y' = g'(f(x)) \times f'(x) \)

In this context, you first determine the derivative of the outer function at the inner function \( g'(f(x)) \) and multiply it by the derivative of the inner function \( f'(x) \). The chain rule helps in finding \( v' \) in our given exercise when \( v = (x^2 + 1)^2 \), by first treating it in the form \( (some \, function)^2 \) and then applying derivatives layer by layer.

Differentiation

Differentiation is the cornerstone of calculus. It's the mathematical way to analyze how a function changes as its input changes. Specifically, it allows you to find the derivative, which is the function giving the rate of change or slope of the original function.

Using rules like the quotient rule, product rule, and chain rule, differentiation becomes a toolkit to solve a wide array of problems, from simple to complex. In the exercise about finding the second derivative, differentiation is used step-by-step:

• First, apply differentiation once using the quotient rule to find \( f'(x) \).
• Next, differentiate \( f'(x) \) again to obtain \( f''(x) \), leveraging both the quotient and chain rules as needed.

The purpose is to explore changes not just of the function itself but how quickly it is changing at different points, which is essential for deeper mathematical analysis, such as understanding acceleration in physics or marginal cost in economics.