Problem 42 - Calculus - AP Edition

Question

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

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)

The function given is \( y = 4x^7 - 5x^3 + 2x \). First, differentiate it with respect to \( x \) to get \( \frac{dy}{dx} \). Using the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \), we get:\[ \frac{dy}{dx} = 28x^6 - 15x^2 + 2. \] Now, differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \):\[ \frac{d^2y}{dx^2} = 168x^5 - 30x. \]

Step 2: Differentiate Part (b)

The function given is \( y = 3x + 2 \). Differentiate it with respect to \( x \) to find \( \frac{dy}{dx} \). Since the derivative of a constant is zero and the derivative of \( 3x \) is simply \( 3 \), we have:\[ \frac{dy}{dx} = 3. \] Differentiating again, any constant derivative is \( 0 \). Hence:\[ \frac{d^2y}{dx^2} = 0. \]

Step 3: Differentiate Part (c)

The function given is \( y = \frac{3x-2}{5x} \). Rewrite it as \( y = \frac{3}{5} - \frac{2}{5x} \). Differentiate this with respect to \( x \):\[ \frac{dy}{dx} = 0 + \frac{2}{5}x^{-2}. \]Simplifying gives \( \frac{dy}{dx} = -\frac{2}{5x^2} \).Differentiate again:\[ \frac{d^2y}{dx^2} = \frac{4}{5}x^{-3} = \frac{4}{5x^3}. \]

Step 4: Differentiate Part (d)

The function given is \( y = (x^3 - 5)(2x + 3) \). First expand the function:\[ y = x^3(2x) + x^3(3) - 5(2x) - 5(3) = 2x^4 + 3x^3 - 10x - 15. \]Differentiate with respect to \( x \) to find \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = 8x^3 + 9x^2 - 10. \] Differentiate this result again to find \( \frac{d^2y}{dx^2} \):\[ \frac{d^2y}{dx^2} = 24x^2 + 18x. \]

Key Concepts

Power Rule

The power rule is a fundamental principle used in calculus to find the derivative of polynomial functions. It states that if you have a term in the form of \( x^n \), its derivative is \( nx^{n-1} \). This rule simplifies the process of differentiation by providing a straightforward way to compute the derivative of algebraic expressions.

For example, consider the function \( y = 4x^7 \), applying the power rule gives:

• Derivative of \( 4x^7 \) is \( 28x^6 \).

By systematically applying the power rule, you can effectively find the derivatives of more complex polynomial functions.

Second Derivative

The second derivative is the derivative of the derivative of a function. It provides additional information about the behavior of the original function, such as its concavity and points of inflection. To compute the second derivative, simply differentiate the first derivative.

For instance, if \( y = 4x^7 - 5x^3 + 2x \), the first derivative is \( \frac{dy}{dx} = 28x^6 - 15x^2 + 2 \). Differentiating this expression once more gives the second derivative:

• \( \frac{d^2y}{dx^2} = 168x^5 - 30x \).

Keep in mind, the second derivative can help determine the curvature of the graph.

Differentiation Steps

Differentiation involves breaking down a function into simpler components. It's helpful to follow a sequence of steps, especially with complex expressions.

Here's a simple sequence to follow:

• Identify each term in the expression.
• Apply the power rule or other appropriate rules to each term.
• Combine the resulting derivatives.

When differentiating, ensure accuracy with each step to maintain the integrity of the final result.

By gradually progressing through each step, you maintain control and clarity over the process.

Simplification

Simplification in calculus often involves rewriting expressions in an easier or alternative form. This makes differentiation and further calculus operations manageable.

Consider the expression \( y = \frac{3x-2}{5x} \). Rewriting it as \( y = \frac{3}{5} - \frac{2}{5x} \) makes the differentiation process more straightforward.

• Simplification reduces complexity.
• It reveals the structure of the function.

Simplify where possible to ease calculations and reduce errors.

Polynomial Functions

Polynomial functions are expressions composed of terms in the form \( ax^n \), where \( a \) is a constant and \( n \) is a non-negative integer. They are one of the basic types of algebraic expressions in calculus.

Characteristics of polynomial functions include:

• Smooth and continuous graphs.
• Defined for all real numbers.
• Relatively simple derivatives using the power rule.

Understanding polynomial functions and their behavior is crucial for solving many calculus problems. They serve as an excellent foundation for learning more complex mathematical concepts.