Problem 33 - Calculus and its applications

Question

Find \(y^{\prime \prime}\) $$ y=\left(x^{3}-2\right)(5 x+1) $$

Step-by-Step Solution

Answer

The second derivative is \( y'' = 60x^2 + 6x \).

Step 1: Understanding the Problem

We are given the function \( y = (x^3 - 2)(5x + 1) \) and need to find its second derivative, \( y'' \). This requires computing the first derivative, \( y' \), and then differentiating again.

Step 2: Expand the Function

First, let's expand \( y = (x^3 - 2)(5x + 1) \) by multiplying the terms: \[ y = x^3(5x + 1) - 2(5x + 1) = 5x^4 + x^3 - 10x - 2 \].

Step 3: Find the First Derivative

Differentiate \( y = 5x^4 + x^3 - 10x - 2 \) with respect to \( x \). Apply the power rule: \[ y' = \frac{d}{dx}(5x^4) + \frac{d}{dx}(x^3) + \frac{d}{dx}(-10x) + \frac{d}{dx}(-2) \]. This gives: \[ y' = 20x^3 + 3x^2 - 10 \].

Step 4: Find the Second Derivative

Now differentiate \( y' = 20x^3 + 3x^2 - 10 \) with respect to \( x \) again: \[ y'' = \frac{d}{dx}(20x^3) + \frac{d}{dx}(3x^2) + \frac{d}{dx}(-10) \]. This results in: \[ y'' = 60x^2 + 6x \].

Key Concepts

Power Rule

The power rule is a simple yet powerful technique used in calculus to differentiate functions. It specifically applies to terms in the form of \( x^n \) where \( n \) is any real number. The rule states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). You can think of this as lowering the power by one and multiplying by the original power.
For example, if you have a term \( 5x^4 \), the derivative is found by applying the power rule: you multiply the term by 4 (because 4 is the original exponent), and then reduce the power by one, resulting in \( 20x^3 \).

• For \( x^3 \), the derivative is \( 3x^2 \).
• For a constant like \( -10 \), its derivative is \( 0 \).

Thus, the power rule is crucial for quickly finding derivatives, especially for polynomial functions where each term can be examined individually and rewritten easily.

Function Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents the rate at which a function's value changes. It is fundamental in calculus and essential for understanding the behavior of functions.
When differentiating a function like \( y = 5x^4 + x^3 - 10x - 2 \), you apply differentiation rules to each term separately. Consistency is crucial. For each term, use the power rule where applicable:

• Differentiate \( 5x^4 \) to get \( 20x^3 \).
• Differentiate \( x^3 \) to get \( 3x^2 \).
• Differentiate \( -10x \) to get \( -10 \).
• Since \(-2\) is constant, it differentiates to \( 0 \).

After applying these rules, the first derivative, \( y' \), tells us how \( y \) changes per unit change in \( x \). Differentiating again gives \( y'' \), the second derivative, indicating the rate of change of the rate of change, which helps us analyze curves and concavity.

Polynomial Expansion

Polynomial expansion is the process of multiplying polynomials to turn a product into a sum. This is often a preliminary step before differentiation, as it transforms the expression into a format easier to manipulate. When dealing with functions like \( y = (x^3 - 2)(5x + 1) \), expanding the product is vital to simplify it for differentiation.
To expand \( y \, \), distribute each term in the first polynomial to every term in the second:

• Multiply \( x^3 \) by \( 5x \) and then by \( 1 \).
• Multiply \( -2 \) by \( 5x \) and then by \( 1 \).

Performing these multiplication steps gives \( y = 5x^4 + x^3 - 10x - 2 \), turning it into a standard polynomial form.
This format allows you to apply the power rule easily to find derivatives. Therefore, polynomial expansion is a handy method for ensuring the effective application of derivative rules to derive necessary functions.