Problem 44 - Calculus Early Transcendentals

Question

Find \(y^{\prime \prime \prime}\) $$ \begin{array}{ll}{\text { (a) } y=5 x^{2}-4 x+7} & {\text { (b) } y=3 x^{-2}+4 x^{-1}+x} \\ {\text { (c) } y=a x^{4}+b x^{2}+c} & {(a, b, c\text { constant) } }\end{array} $$

Step-by-Step Solution

Answer

(a) 0, (b) \(-72x^{-5} - 24x^{-4}\), (c) \(24ax\)

Step 1: Differentiate Part (a) - First Derivative

Given the function \( y = 5x^2 - 4x + 7 \), find the first derivative. Differentiate term by term:- The derivative of \( 5x^2 \) is \( 10x \).- The derivative of \( -4x \) is \( -4 \).- The derivative of the constant 7 is 0.Thus, the first derivative \( y' = 10x - 4 \).

Step 2: Differentiate Part (a) - Second and Third Derivative

Now find the second and third derivatives:- The second derivative \( y'' \) of \( 10x - 4 \) is 10.- The third derivative \( y''' \) is the derivative of 10, which is 0.

Step 3: Differentiate Part (b) - First Derivative

Given \( y = 3x^{-2} + 4x^{-1} + x \), find the first derivative:- The derivative of \( 3x^{-2} \) is \( -6x^{-3} \).- The derivative of \( 4x^{-1} \) is \( -4x^{-2} \).- The derivative of \( x \) is 1.Therefore, the first derivative \( y' = -6x^{-3} - 4x^{-2} + 1 \).

Step 4: Differentiate Part (b) - Second and Third Derivative

Find the second and third derivatives:- The second derivative \( y'' \) is the derivative of \( -6x^{-3} - 4x^{-2} + 1 \), which is \( 18x^{-4} + 8x^{-3} \).- The third derivative \( y''' \) is the derivative of \( 18x^{-4} + 8x^{-3} \), resulting in \( -72x^{-5} - 24x^{-4} \).

Step 5: Differentiate Part (c) - First Derivative

Given \( y = ax^4 + bx^2 + c \), differentiate to find \( y' \):- The derivative of \( ax^4 \) is \( 4ax^3 \).- The derivative of \( bx^2 \) is \( 2bx \).- The derivative of the constant \( c \) is 0.Thus, the first derivative \( y' = 4ax^3 + 2bx \).

Step 6: Differentiate Part (c) - Second and Third Derivative

Find the second and third derivatives:- The second derivative \( y'' \) is the derivative of \( 4ax^3 + 2bx \), which is \( 12ax^2 + 2b \).- The third derivative \( y''' \) is the derivative of \( 12ax^2 + 2b \), resulting in \( 24ax \).

Key Concepts

First Derivative

The first derivative of a function is like finding the slope of the curve it represents. If we have a function, such as a simple polynomial like in Part (a) with the expression \( y = 5x^2 - 4x + 7 \), the first derivative tells us how \( y \) changes as \( x \) changes. In simple terms, it measures the rate of change.

• For the term \( 5x^2 \), the derivative becomes \( 10x \).
• The term \( -4x \) becomes \( -4 \).
• The constant \( 7 \) simply disappears since constants don't change, represented by a derivative of \( 0 \).

Thus, the first derivative for Part (a) is \( y' = 10x - 4 \). Each step shows how to handle different types of terms: powers, constants, and coefficients.

Second Derivative

The second derivative is the derivative of the first derivative. In essence, it measures the rate at which the rate of change is changing. This concept is crucial in identifying concavity and intervals of increase or decrease in a function.For Part (a), from the first derivative \( y' = 10x - 4 \):

• The derivative of \( 10x \) is \( 10 \).
• The derivative of \( -4 \) (a constant) is \( 0 \).

So, the second derivative is \( y'' = 10 \), indicating a constant rate of change.In Part (b), the first derivative is more complex: \( y' = -6x^{-3} - 4x^{-2} + 1 \). Applying the derivative rules here:

• \( -6x^{-3} \) becomes \( 18x^{-4} \).
• \( -4x^{-2} \) becomes \( 8x^{-3} \).
• The constant \( 1 \) becomes \( 0 \).

This results in the second derivative \( y'' = 18x^{-4} + 8x^{-3} \).

Third Derivative

The third derivative is the derivative of the second derivative. It provides insight into how the concavity itself is changing and is used in advancement problems involving motion, waves, and more.
Let's look at Part (b) again, where the second derivative is \( y'' = 18x^{-4} + 8x^{-3} \):

• \( 18x^{-4} \) differentiates to \( -72x^{-5} \).
• \( 8x^{-3} \) differentiates to \( -24x^{-4} \).

Thus, the third derivative becomes \( y''' = -72x^{-5} - 24x^{-4} \).For Part (c), simplifying again shows that taking a derivative reduces polynomial powers and their complexity until a zero constant remains, as seen with \( y''' = 24ax \) from \( 12ax^2 + 2b \).
Understanding third derivatives enhances comprehension of how dynamic systems change at deeper levels and assists in precision modeling of physical processes.