Problem 58 - Single Variable Calculus

Question

Find \(y^{\prime \prime}\). \(y=\frac{x}{x^{2}+1}\)

Step-by-Step Solution

Answer

The second derivative of the given function is: \(y^{\prime \prime} = \frac{2x^{5} + 2x^{3} - 2x}{(x^{2} + 1)^{4}}\).

Step 1: Apply the quotient rule to find the first derivative

Recall the quotient rule: If the function \(g(x) = \frac{f(x)}{h(x)}\), then \(g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^{2}}\). Here, \(f(x) = x\) and \(h(x) = x^{2} + 1\). Note that \(f'(x) = 1\) and \(h'(x) = 2x\). Applying the quotient rule to the given function, we have: \(y' = \frac{1(x^2 + 1) - x(2x)}{(x^{2} + 1)^{2}}\) Now simplify the numerator: \(y' = \frac{x^2 + 1 - 2x^2}{(x^{2} + 1)^{2}}\)

Step 2: Simplify the first derivative

Combine the terms in the numerator: \(y' = \frac{-x^2 + 1}{(x^{2} + 1)^{2}}\) Now we have the first derivative of the given function.

Step 3: Apply the quotient rule to find the second derivative

Now to find the second derivative, we will once again apply the quotient rule to the simplified first derivative. Let \(f(x) = -x^2 + 1\) and \(h(x) = (x^{2} + 1)^{2}\). Note that \(f'(x) = -2x\) and \(h'(x) = 4x(x^{2} + 1)\) (using the chain rule for the second term). Then, applying the quotient rule to find the second derivative: \(y'' = \frac{(-2x)((x^{2} + 1)^{2}) - (-x^{2} + 1)4x(x^{2} + 1)}{(x^{2} + 1)^{4}}\)

Step 4: Simplify the second derivative

Now we need to simplify the numerator of the expression for the second derivative: \(y'' = \frac{-2x(x^{4} + 2x^{2} + 1) + 4x^2(x^{3} + x)}{(x^{2} + 1)^{4}}\) Expand and simplify the numerator: \(y'' = \frac{-2x^{5} - 4x^{3} - 2x + 4x^{5} + 4x^{3}}{(x^{2} + 1)^{4}}\) Combine like terms: \(y'' = \frac{2x^{5} + 2x^{3} - 2x}{(x^{2} + 1)^{4}}\) The second derivative of the given function is: \(y^{\prime \prime} = \frac{2x^{5} + 2x^{3} - 2x}{(x^{2} + 1)^{4}}\)

Key Concepts

Quotient Rule Differentiation

Understanding the quotient rule is essential when differentiating functions that are presented as ratios of two other functions. The quotient rule formula is \[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^{2}}\]When we differentiate a quotient, we're essentially finding the rate at which the ratio of the two functions is changing. The numerator involves a difference representing the change in the top function, scaled by the value of the bottom function, minus the change in the bottom function scaled by the value of the top function. It is crucial that students apply this rule methodically as improper application can easily lead to miscalculations.

Exercise Application

In our exercise \(y=\frac{x}{x^{2}+1}\), the top function \(f(x)\) and the bottom function \(g(x)\) have been differentiated separately as \(f'(x)\) and \(g'(x)\), and then applied to the quotient rule formula. This methodical approach must be followed carefully to ensure accuracy, particularly as functions become more complex.

First Derivative Calculation

The first derivative of a function gives us the slope of the tangent line to the curve of the function at any point. It is a fundamental concept in calculus that signifies the rate of change or the velocity of change of the function's value. Calculating the first derivative often involves direct differentiation using basic rules, such as the power rule, sum rule, and product rule, as well as more complex ones like the quotient and chain rules for more intricate functions. The first derivative can also reveal important characteristics about the function, such as increasing or decreasing behavior and potential local maxima or minima.

Exercise Insight

In our exercise, we calculated the first derivative by applying the quotient rule. This step is vital before proceeding to the second derivative, and simplifying expressions—as we did by combining like terms—can make following steps less cumbersome. It's equally important to review the derived expression to understand the behavior of the function.

Chain Rule Application

The chain rule is a rule in calculus for differentiating composite functions. When a function is composed of an outer function \( g(x) \) and an inner function \( f(x) \), the derivative of this composite function is given by the product of the derivative of the outer function evaluated at the inner function, and the derivative of the inner function. The chain rule formula is \[\frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x)\]The chain rule allows us to unravel the layers of complexity in functions that are nested within each other. When students tackle such functions without applying the chain rule correctly, they risk producing erroneous results.

Exercise Application

The second derivative in our example required the application of the chain rule to find \( h'(x) \) since \( h(x) \) is a function of a function—\( (x^{2} + 1)^{2} \). Mastery of the chain rule is critical because it is referenced whenever derivatives of composite functions are calculated, such as in the second derivative of our problem.