Problem 31 - Brief Applied Calculus

Question

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

Step-by-Step Solution

Answer

The second derivative is \(\frac{16}{(2x+1)^3}\).

Step 1: Find the First Derivative

To find the first derivative of the given function, use the quotient rule, which states ifg(y) = u(x)/v(x), then the derivative g'(x) is given by \( g'(x) = \frac{u'(x)v(x)-u(x)v'(x)}{(v(x))^2} \). In this function, \( u(x) = 2x - 1 \) and \( v(x) = 2x + 1 \). Find the derivatives:\( u'(x) = 2 \) and \( v'(x) = 2 \).Apply the quotient rule:\[ \frac{d}{dx} \left( \frac{2x-1}{2x+1} \right) = \frac{(2)(2x+1) - (2x-1)(2)}{(2x+1)^2} \].Simplify:\[ \frac{4x + 2 - 4x + 2}{(2x+1)^2} = \frac{4}{(2x+1)^2} \].Thus, the first derivative is \( f'(x) = \frac{4}{(2x+1)^2} \).

Step 2: Set Up Derivative for the Second Derivative

To find the second derivative, we need to differentiate \( f'(x) = \frac{4}{(2x+1)^2} \) once again. We can use the chain rule to find the derivative of a function in the form of \( g(x) = \frac{c}{u(x)^n} \), where \( g'(x) = -cn \frac{u'(x)}{u(x)^{n+1}} \).Identify \( c = 4 \), \( u(x) = 2x+1 \), \( n = -2 \), and \( u'(x) = 2 \).

Step 3: Differentiate to Find Second Derivative

To differentiate \( f'(x) = \frac{4}{(2x+1)^2} \), apply the chain rule:\[ f''(x) = -4(-2) \cdot 2 \cdot (2x+1)^{-3} \].Simplify:\[ = 16 \cdot (2x+1)^{-3} \].Rewrite it in a familiar form:\[ = \frac{16}{(2x+1)^3} \].Thus, the second derivative \( f''(x) = \frac{16}{(2x+1)^3} \).

Key Concepts

Quotient Rule

The Quotient Rule is an essential technique in calculus used to differentiate functions that are expressed as the division of two sub-functions. If we have a function expressed as \( g(x) = \frac{u(x)}{v(x)} \), the Quotient Rule helps us find its derivative. It states that

• The derivative, \( g'(x) \), is given by \( \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2} \).

To understand this better, in our problem, the function \( f(x) = \frac{2x-1}{2x+1} \) requires us to find the derivative using the Quotient Rule.

• Here, \( u(x) = 2x - 1 \) and its derivative, \( u'(x) = 2 \).
• The denominator function, \( v(x) = 2x + 1 \), has a derivative of \( v'(x) = 2 \).

By applying the formula, the first derivative becomes \( f'(x) = \frac{(2)(2x+1) - (2x-1)(2)}{(2x+1)^2} \), which simplifies to \( \frac{4}{(2x+1)^2} \). This approach is vital for precisely breaking down complex ratios and arriving at correct derivatives.

Chain Rule

The Chain Rule applies when differentiating compositions of functions, a frequent scenario in calculus. Essentially, it defines how to differentiate a function \( h(x) \) composed of an outer function and an inner function, expressed as \( f(g(x)) \).

• The Chain Rule states: if \( y = f(g(x)) \), then \( y' = f'(g(x)) \cdot g'(x) \).

In our original problem, the expression for the second derivative involves the function \( f'(x) = \frac{4}{(2x+1)^2} \), which can be considered in the form \( c \cdot u(x)^{-n} \).
To find the second derivative, we need to differentiate \( u(x) = 2x+1 \) and incorporate it.

• We apply the Chain Rule: \( f''(x) = -4(-2) \cdot u'(x) \cdot (2x+1)^{-3} \), simplifying to \( \frac{16}{(2x+1)^3} \).
• This simplifies the derivative process when a function has nested layers, allowing us to manage each separately and combine them effectively.

Calculus Differentiation

Calculus Differentiation encompasses all techniques used to find the rate at which a function changes. It is the heart of calculus and provides tools to uncover the underlying behavior of mathematical models by analyzing slopes of curves. Differentiating means finding the derivative, a measure of how a function's output changes as its input changes.
In our original exercise, we are tasked with finding both first and second derivatives for the function \( f(x) = \frac{2x-1}{2x+1} \). Calculus differentiation involves several steps and rules, including:

• Applying the Quotient Rule to find the first derivative: this is crucial for functions expressed as fractions.
• Using the Chain Rule for further differentiation, especially when functions involve complex compositions or indices.

Differentiation reveals the behavior and features of functions, such as rates of change at particular points, maxima, minima, and inference of curvature. Mastering differentiation is crucial for delving deeper into calculus and solving real-world problems involving rates of change and motion.