Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(1-8\), find \(d^{3} y / d x^{3}\). $$ y=\frac{3 x}{1-x} $$

Step-by-Step Solution

Answer

The third derivative \( \frac{d^3y}{dx^3} = \frac{18}{(1-x)^4} \).

Step 1: Find the First Derivative

To find the first derivative \( \frac{dy}{dx} \) of \( y = \frac{3x}{1-x} \), apply the quotient rule: \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = 3x \) and \( v = 1-x \). The derivatives are \( u' = 3 \) and \( v' = -1 \). Thus, \( \frac{dy}{dx} = \frac{3(1-x) - 3x(-1)}{(1-x)^2} = \frac{3 - 3x + 3x}{(1-x)^2} = \frac{3}{(1-x)^2} \).

Step 2: Find the Second Derivative

Differentiate the first derivative \( \frac{dy}{dx} = \frac{3}{(1-x)^2} \) with respect to \( x \). Using the chain rule, estimate the derivative: \( \frac{d}{dx}[ (1-x)^{-2} ] = -2(1-x)^{-3}(-1) \). So, \( \frac{d^2y}{dx^2} = 3 \times 2(1-x)^{-3} = \frac{6}{(1-x)^3} \).

Step 3: Find the Third Derivative

Differentiate the second derivative \( \frac{d^2y}{dx^2} = \frac{6}{(1-x)^3} \). Using the chain rule again, \( \frac{d}{dx}[ (1-x)^{-3} ] = -3(1-x)^{-4}(-1) \). Thus, \( \frac{d^3y}{dx^3} = 6 \times 3(1-x)^{-4} = \frac{18}{(1-x)^4} \).

Key Concepts

Quotient Rule

The quotient rule is a technique used in calculus when you need to differentiate a function that is the division of two other functions. For a function given by the quotient \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule states that the derivative \( \frac{dy}{dx} \) is calculated as:

• Numerator: \( u'v - uv' \), where \( u' \) is the derivative of \( u \) and \( v' \) is the derivative of \( v \).
• Denominator: \( v^2 \), which is the square of the function \( v \).

This results in the formula: \[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]Using this rule allows us to find the first derivative of a given quotient function. In our exercise, for example, with \( u = 3x \) and \( v = 1-x \), applying the quotient rule gives us the first derivative as a simplified function. Remember that careful handling of negative signs in the derivatives is crucial to getting the right result.

Chain Rule

To differentiate composite functions, where one function is applied to the result of another, the chain rule is essential. It is used when a function exists within the domain of another function, forming a composition like \( f(g(x)) \). The chain rule formula is:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In terms of our problem, the function inside the derivative takes the form \((1-x)^{-n}\). Using the chain rule means we differentiate the outer function as normal, and then multiply by the derivative of the inner function \((1-x)\). This step is crucial in finding both the second and third derivatives in our problem example. Pay attention to the power rule, as well, since it affects the exponent when differentiating.

Differentiation Steps

Differentiating functions involves a series of carefully executed steps, especially when finding higher-order derivatives like the third derivative \( \frac{d^3y}{dx^3} \).

• First Derivative: Use initial rules like the quotient rule to find the first derivative, simplifying expressions wherever possible.
• Second Derivative: Apply the chain rule, focusing on differentiating the rational function repeatedly as needed.
• Third Derivative: Continue using the chain rule and adjust coefficients accordingly, being careful with negative signs and changes in exponents.

Each derivative builds upon the previous one. Success in finding the third derivative, as seen in our exercise, hinges on correctly applying the rules and simplifying steps at each stage. The practice significantly strengthens understanding of calculus basics and progress towards more complex equations.