Problem 53 - Calculus Early Transcendentals

Question

Find \(d^{2} y / d x^{2}\) $$ y=\frac{1+x}{1-x} $$

Step-by-Step Solution

Answer

The second derivative is \( \frac{4}{(1-x)^3} \).

Step 1: Find the First Derivative

To find the second derivative, we first need the first derivative of the function. We'll use the quotient rule for differentiation: if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Here, \( u = 1 + x \) and \( v = 1 - x \). First, we find \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -1 \). Substitute these into the quotient rule formula: \[ \frac{dy}{dx} = \frac{(1-x) \cdot 1 - (1+x) \cdot (-1)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2} \].

Step 2: Differentiate Again for the Second Derivative

Now, we differentiate \( \frac{dy}{dx} = \frac{2}{(1-x)^2} \) to find \( \frac{d^2 y}{dx^2} \). This requires the chain rule. Rewrite the function as \( 2(1-x)^{-2} \). The derivative of \( (1-x)^{-2} \) is \(-2(1-x)^{-3} \cdot (-1)\). Apply the chain rule: \( \frac{d}{dx}[2(1-x)^{-2}] = 2 \cdot -2 \cdot (1-x)^{-3} \cdot (-1) = \frac{4}{(1-x)^3} \).

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus used to differentiate functions that are in the form of a fraction. Consider a function given by the ratio \( y = \frac{u}{v} \), where \( u \) and \( v \) are differentiable functions of \( x \). The derivative of this function is found through the formula:

• \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

This rule combines the derivatives of the numerator \( u \) and the denominator \( v \) separately, taking into account their roles in the fraction. In the exercise, \( u = 1 + x \) and \( v = 1 - x \). When using the Quotient Rule:

• The derivative of \( u \) with respect to \( x \) is 1.
• The derivative of \( v \) with respect to \( x \) is -1.

Substituting these into the formula, we focused on simplifying the expression to attain the first derivative as \( \frac{dy}{dx} = \frac{2}{(1-x)^2} \). Using the Quotient Rule properly ensures precision in deriving functions expressed as quotients.

Chain Rule

The Chain Rule is a key method used in calculus for differentiating composite functions. It is applied when a function is composed of other functions. Given a situation where \( y = f(g(x)) \), the derivative of \( y \) with respect to \( x \) is calculated by:

• Finding the derivative of the outer function, \( f' \), evaluated at the inner function \( g(x) \).
• Multiplying by the derivative of the inner function, \( g'(x) \).

In the exercise, when calculating the second derivative, we transformed the first derivative \( \frac{2}{(1-x)^2} \) into a form suitable for the Chain Rule. By rewriting it as \( 2(1-x)^{-2} \), the differentiation became clearer:

• The outer function is \( 2x^{-2} \), and its derivative is \( -2x^{-3} \).
• The inner function is \( 1-x \), and its derivative is \( -1 \).

Together, these steps illustrate how to handle nested functions by carefully addressing both their outer and inner components, resulting in \( \frac{d^2 y}{dx^2} = \frac{4}{(1-x)^3} \). The power of the Chain Rule becomes apparent when dealing with such layered expressions.

Calculus

Calculus is a branch of mathematics that deals with continuous change, through the concepts of derivatives and integrals. It provides essential tools for analyzing the rate of change and accumulation processes, applicable in various fields such as physics, engineering, economics, and beyond.
The exercise illustrates a common application: finding the second derivative of a function. To perform this task, calculus employs differentiation techniques like the Quotient Rule and Chain Rule. These principles are foundational for understanding how functions evolve, allowing us to quantify precise changes in contexts where everything is in motion.
Derivatives, as demonstrated, help us deduce the behavior of functions by providing instant rates of change. The practice of finding the second derivative, in particular, unfolds deeper insights, such as concavity and acceleration, which further enrich the predictive power of calculus. Solving the exercise with a systematic approach not only applies calculus but also showcases its elegance and utility.