Problem 42 - Calculus Early Transcendentals

Question

Find \(d^{2} y / d x^{2}\) $$ \begin{array}{ll}{\text { (a) } y=4 x^{7}-5 x^{3}+2 x} & {\text { (b) } y=3 x+2} \\ {\text { (c) } y=\frac{3 x-2}{5 x}} & {\text { (d) } y=\left(x^{3}-5\right)(2 x+3)}\end{array} $$

Step-by-Step Solution

Answer

(a) \( 168x^5 - 30x \); (b) 0; (c) \( -\frac{4}{5x^3} \); (d) \( 24x^2 + 18x \)."

Step 1: Differentiate Part (a)

Given the function \( y = 4x^7 - 5x^3 + 2x \), differentiate with respect to \( x \):1. The derivative of \( 4x^7 \) is \( 28x^6 \).2. The derivative of \( -5x^3 \) is \( -15x^2 \).3. The derivative of \( 2x \) is \( 2 \).Thus, the first derivative \( \frac{dy}{dx} = 28x^6 - 15x^2 + 2 \).Next, differentiate again to find \( \frac{d^2y}{dx^2} \):1. The derivative of \( 28x^6 \) is \( 168x^5 \).2. The derivative of \( -15x^2 \) is \( -30x \).3. The derivative of \( 2 \) is \( 0 \).Thus, \( \frac{d^2y}{dx^2} = 168x^5 - 30x \).

Step 2: Differentiate Part (b)

Given the function \( y = 3x + 2 \), differentiate with respect to \( x \):1. The derivative of \( 3x \) is \( 3 \).2. The derivative of \( 2 \) is \( 0 \).Thus, the first derivative is \( \frac{dy}{dx} = 3 \).For the second derivative, differentiate \( 3 \), which is constant:Thus, \( \frac{d^2y}{dx^2} = 0 \).

Step 3: Differentiate Part (c) using Quotient Rule

Given the function \( y = \frac{3x - 2}{5x} \), use the Quotient Rule:\[ \frac{dy}{dx} = \frac{(5x)(3) - (3x-2)(5)}{(5x)^2} = \frac{15x - (15x - 10)}{25x^2} = \frac{10}{25x^2} = \frac{2}{5x^2} \].Differentiate again using the chain rule:\( \frac{d}{dx}\left(\frac{2}{5x^2}\right) = -\frac{4}{5x^3} \).Thus, \( \frac{d^2y}{dx^2} = -\frac{4}{5x^3} \).

Step 4: Differentiate Part (d) using Product Rule

Given the function \( y = (x^3 - 5)(2x + 3) \), expand first:\[ y = 2x^4 + 3x^3 - 10x - 15 \].Differentiate with respect to \( x \) using the sum:1. The derivative of \( 2x^4 \) is \( 8x^3 \).2. The derivative of \( 3x^3 \) is \( 9x^2 \).3. The derivative of \( -10x \) is \( -10 \).4. The derivative of \( -15 \) is \( 0 \).Thus, \( \frac{dy}{dx} = 8x^3 + 9x^2 - 10 \).Differentiate again:1. The derivative of \( 8x^3 \) is \( 24x^2 \).2. The derivative of \( 9x^2 \) is \( 18x \).3. The derivative of \( -10 \) is \( 0 \).Thus, \( \frac{d^2y}{dx^2} = 24x^2 + 18x \).

Key Concepts

Differentiation

Differentiation is a powerful mathematical tool that allows us to find the rate at which a quantity changes. It is fundamental in calculus and helps determine the slope of a curve at any given point. Think of it like zooming in on a function to see how steeply it rises or falls. Differentiation involves finding the derivative of a function, which is the result of applying specific rules to calculate this rate of change. There are several rules of differentiation, which make the process of finding a derivative very efficient, such as the power rule, product rule, quotient rule, and chain rule. Through differentiation, you are essentially peeling back layers to uncover how a function behaves, providing essential insights in both theoretical and applied mathematics.

In practice, if you have a function like \( y = 4x^7 - 5x^3 + 2x \), you can differentiate it to find \( \frac{dy}{dx} \), and by differentiating again, you can discover the second derivative, \( \frac{d^2y}{dx^2} \), revealing the function's concavity.

Quotient Rule

The quotient rule is utilized when differentiating functions that are fractions of two expressions. It is particularly handy when you're dealing with a division where both the numerator and the denominator are functions of the variable you're differentiating. The quotient rule formula is as follows:

• If you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative \( \frac{dy}{dx} \) is given by: \( \frac{v(du/dx) - u(dv/dx)}{v^2} \).

This formula highlights the need to differentiate the numerator and denominator separately, subtracting the product of the denominator and the derivative of the numerator from the product of the numerator and the derivative of the denominator.

For example, if applied to the function \( y = \frac{3x - 2}{5x} \), using the quotient rule would yield the first derivative \( \frac{2}{5x^2} \). Applying the chain rule afterwards helps us find the second derivative.

Product Rule

The product rule is a basic differentiation rule used when dealing with the multiplication of two functions. When you have a function expressed as a product, the product rule helps you find the derivative without expanding it first. The formula for the product rule is:

• For a function \( y = u \cdot v \), the derivative \( \frac{dy}{dx} \) is: \( u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \).

This means you take the derivative of each function separately, multiplying each by the other original function, and then sum the results.

In the given exercise, \( y = (x^3 - 5)(2x + 3) \) was simplified first; however, using the product rule directly would also have been an efficient method to find the first derivative. The second derivative involves differentiating the first derivative, using basic differentiation techniques.

Chain Rule

The chain rule is essential when you're faced with composite functions, where one function is nested within another. This rule allows you to differentiate these complex functions efficiently. The chain rule is articulated as:

• If \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is the derivative of \( f \) with respect to \( g \), times the derivative of \( g \) with respect to \( x \): \( f'(g(x)) \cdot g'(x) \).

This rule is like unwrapping layers, first differentiating the outer function, and then multiplying it by the derivative of the inner function.

In the exercise example of \( y = \frac{3x - 2}{5x} \), after finding the first derivative, the chain rule was applied to find the second derivative, particularly while dealing with expressions involving powers of a function of \( x \). Using the chain rule ensures you are accounting for all parts of these nested functions.