Problem 41 - Calculus Early Transcendentals

Question

Find \(d^{2} y / d x^{2}\) $$ \begin{array}{ll}{\text { (a) } y=7 x^{3}-5 x^{2}+x} & {\text { (b) } y=12 x^{2}-2 x+3} \\ {\text { (c) } y=\frac{x+1}{x}} & {\text { (d) } y=\left(5 x^{2}-3\right)\left(7 x^{3}+x\right)}\end{array} $$

Step-by-Step Solution

Answer

(a) 42x - 10, (b) 24, (c) 2/x^3, (d) 700x^3 - 96x

Step 1a: Differentiate the first equation (first derivative)

Given the function \( y = 7x^3 - 5x^2 + x \), calculate the first derivative \( \frac{dy}{dx} \):- Differentiate \( 7x^3 \) to get \( 21x^2 \)- Differentiate \( -5x^2 \) to get \( -10x \)- Differentiate \( x \) to get \( 1 \)The first derivative is \( \frac{dy}{dx} = 21x^2 - 10x + 1 \).

Step 2a: Differentiate again (second derivative)

To find \( \frac{d^2y}{dx^2} \) for the function \( y = 7x^3 - 5x^2 + x \), differentiate \( \frac{dy}{dx} = 21x^2 - 10x + 1 \):- Differentiate \( 21x^2 \) to get \( 42x \)- Differentiate \( -10x \) to get \( -10 \)- The derivative of \( 1 \) is \( 0 \)The second derivative is \( \frac{d^2y}{dx^2} = 42x - 10 \).

Step 1b: Differentiate the second equation (first derivative)

Given the function \( y = 12x^2 - 2x + 3 \), compute the first derivative:- Differentiate \( 12x^2 \) to get \( 24x \)- Differentiate \( -2x \) to get \( -2 \)- The derivative of the constant \( 3 \) is \( 0 \)The first derivative is \( \frac{dy}{dx} = 24x - 2 \).

Step 2b: Differentiate again (second derivative)

Find \( \frac{d^2y}{dx^2} \) for \( y = 12x^2 - 2x + 3 \):- Differentiate \( 24x \) to get \( 24 \)- The derivative of \( -2 \) is \( 0 \)Thus, the second derivative is \( \frac{d^2y}{dx^2} = 24 \).

Step 1c: Rewrite the third function for differentiation

Given \( y = \frac{x+1}{x} \), rewrite as \( y = 1 + \frac{1}{x} \) or \( y = 1 + x^{-1} \).

Step 2c: Differentiate the rewritten function (first derivative)

Compute the first derivative of \( y = 1 + x^{-1} \):- The derivative of \( 1 \) is \( 0 \)- Differentiate \( x^{-1} \) to get \( -x^{-2} \) or \( -\frac{1}{x^2} \)The first derivative is \( \frac{dy}{dx} = -x^{-2} \).

Step 3c: Differentiate again (second derivative)

Find \( \frac{d^2y}{dx^2} \) for \( y = 1 + x^{-1} \):- Differentiate \( -x^{-2} \) to get \( 2x^{-3} \) or \( \frac{2}{x^3} \)The second derivative is \( \frac{d^2y}{dx^2} = \frac{2}{x^3} \).

Step 1d: Expand the product for differentiation

Given \( y = (5x^2 - 3)(7x^3 + x) \), expand the expression:- First, use the distributive property: \[ y = 5x^2 \cdot 7x^3 + 5x^2 \cdot x - 3 \cdot 7x^3 - 3 \cdot x \]- Simplify: \[ y = 35x^5 + 5x^3 - 21x^3 - 3x \]- Combine like terms: \[ y = 35x^5 - 16x^3 - 3x \]

Step 2d: Differentiate the expanded function (first derivative)

Compute the first derivative:- Differentiate \( 35x^5 \) to get \( 175x^4 \)- Differentiate \( -16x^3 \) to get \( -48x^2 \)- Differentiate \( -3x \) to get \( -3 \)The first derivative is \( \frac{dy}{dx} = 175x^4 - 48x^2 - 3 \).

Step 3d: Differentiate again (second derivative)

Find \( \frac{d^2y}{dx^2} \) for the expanded function:- Differentiate \( 175x^4 \) to get \( 700x^3 \)- Differentiate \( -48x^2 \) to get \( -96x \)- The derivative of \( -3 \) is \( 0 \)The second derivative is \( \frac{d^2y}{dx^2} = 700x^3 - 96x \).

Key Concepts

Calculus

Calculus is a fascinating branch of mathematics that deals with change. Imagine tracking a moving car and needing to know how fast it's going – that's where calculus comes into play. It involves two main ideas: differentiation and integration. Differentiation focuses on finding rates of change, like speed, while integration sums up quantities over intervals, like finding the total distance traveled. In this exercise, we're looking at differentiation and particularly the second derivative, which tells us how the slope or rate of change is itself changing.

With each step of differentiation, we're examining the function more deeply:

• The first derivative of a function gives the slope of the tangent line, indicating how fast or slow something is changing.
• The second derivative then tells us about the concavity of the function, essentially pointing out if the change is speeding up or slowing down.

This is immensely useful across a variety of fields like physics, economics, and biology, providing insights into the behavior of changing systems.

Derivative Rules

When dealing with differentiation, we utilize specific derivative rules to make the process smooth and effective. Remembering these rules is like having a toolkit for various functions we encounter:

• Power Rule: For any function of the form \(x^n\), the derivative is \(nx^{n-1}\). This is the rule we applied consistently in the given exercise.
• Sum Rule: When differentiating two or more functions combined by addition or subtraction, differentiate each one separately.
• Quotient and Product Rules: More complex rules that apply to functions divided by, or multiplied with each other. They require careful application to find the derivative accurately.

Using these rules, we systematically find the first and second derivatives, making it possible to understand much more about a function's behavior. By breaking down functions step by step, differentiation becomes manageable and predictable.

Polynomial Differentiation

Polynomials are a common type of function in calculus characterized by the sum of terms of the form \(ax^n\). They are straightforward to differentiate, thanks to their simple structure. Let’s dive into the process with examples from the exercise:

For any polynomial, we can apply the power rule to find the derivative:

• In a function like \(7x^3 - 5x^2 + x\), start by applying the power rule separately to each term.
• The first derivative gives \(21x^2 - 10x + 1\), showing the slope's behavior at each point.
• Differentiating once more, we find the second derivative, revealing the rate of change of the slope itself: \(42x - 10\).

This process is repeated for each function, whether it’s a simple polynomial or a more complex product of polynomials. By focusing on each term individually and using the power rule, polynomial differentiation becomes a straightforward task that unveils detailed insights into the nature of these functions. Whether the result aids in graphing curves or understanding dynamic systems, polynomial differentiation is a powerful tool in mathematical analysis.