Problem 56 - Brief Applied Calculus

Question

Find the second derivative of each function. $$ f(x)=\left(x^{3}-1\right)^{5} $$

Step-by-Step Solution

Answer

The second derivative is \( f''(x) = 30x(x^3 - 1)^4 + 180x^4(x^3 - 1)^3 \).

Step 1: Apply the Chain Rule to Find the First Derivative

We need to find the first derivative of \( f(x) = (x^3 - 1)^5 \). Identify the outer function as \( u^5 \) where \( u = x^3 - 1 \), and the inner function as \( x^3 - 1 \). Using the chain rule, the derivative of \( f(x) \) is \( 5u^4 \cdot \frac{du}{dx} \).

Step 2: Find the Derivative of the Inner Function

The inner function is \( u = x^3 - 1 \). The derivative of this with respect to \( x \) is \( \frac{du}{dx} = 3x^2 \).

Step 3: Combine to Find the First Derivative

Substitute the derivative of the inner function into the derivative found using the chain rule: \( f'(x) = 5(x^3 - 1)^4 \cdot 3x^2 = 15x^2(x^3 - 1)^4 \).

Step 4: Apply the Product Rule for the Second Derivative

To find the second derivative, apply the product rule to \( f'(x) = 15x^2(x^3 - 1)^4 \). Let \( g(x) = 15x^2 \) and \( h(x) = (x^3 - 1)^4 \). The product rule states \( f''(x) = g'(x)h(x) + g(x)h'(x) \).

Step 5: Derive the Components for the Product Rule

Find \( g'(x) = 30x \). Use the chain rule on \( h(x) = (x^3 - 1)^4 \). Let \( u = x^3 - 1 \), \( h'(x) = 4u^3 \cdot 3x^2 = 12x^2(x^3 - 1)^3 \).

Step 6: Combine to Find the Second Derivative

Substitute the derivatives of \( g(x) \) and \( h(x) \) into the product rule: \( f''(x) = 30x(x^3 - 1)^4 + 15x^2 \cdot 12x^2(x^3 - 1)^3 \). Simplify: \( f''(x) = 30x(x^3 - 1)^4 + 180x^4(x^3 - 1)^3 \).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential tool when it comes to differentiation, particularly for composite functions. A composite function is a function made up of two or more other functions. For example, if function \( f(x) \) is composed of an outer function \( u^n \) and an inner function \( u(x) \), the chain rule helps us find the derivative of this function.
The chain rule states: if \( y = f(u) \) and \( u = g(x) \), then the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

• First, find the derivative of the outer function with respect to the inner function.
• Next, find the derivative of the inner function with respect to \( x \).
• Finally, multiply these two derivatives together to get the derivative of the composite function.

In our original exercise, the chain rule was used to find the first derivative of the function \( f(x) = (x^3 - 1)^5 \). By identifying the inner function \( u = x^3 - 1 \) and the outer function \( u^5 \), we were able to successfully apply the chain rule.

Exploring the Product Rule

The product rule is another differentiation technique used when dealing with products of two functions. It becomes handy when we need to find derivatives of functions that are expressed as the product of two simpler functions.
According to the product rule: if you have two functions \( g(x) \) and \( h(x) \), then the derivative of their product \( f(x) = g(x) \cdot h(x) \) is given by:\[ f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) \].

• First, take the derivative of the first function \( g(x) \) and multiply it by the second function \( h(x) \).
• Then, take the derivative of \( h(x) \) and multiply it by \( g(x) \).
• Finally, add these two products together.

In practice, this is exactly what was done in step 4 of the original exercise, where the functions \( g(x) = 15x^2 \) and \( h(x) = (x^3 - 1)^4 \) were used to find the second derivative of \( f(x) \). This ensured all aspects of each function's behavior were captured.

Grasping the Basics of Derivatives

Taking derivatives is a fundamental concept in calculus, used to determine the rate at which one variable changes with respect to another. The derivative measures how a function changes as its input changes, providing the slope of the tangent line to the function at any point.
The process of finding derivatives involves several rules depending on the type of function being differentiated. Some of these include:

• The power rule for functions of the form \( x^n \), where the derivative is \( nx^{n-1} \).
• The chain rule for composite functions, as previously discussed.
• The product rule for products of functions, which was employed in step 4 for finding the second derivative.

When finding the second derivative, we differentiate the first derivative. This provides us with insights into the curvature of the function or its concavity. In the exercise, calculating \( f''(x) \) involved applying both the chain rule and the product rule, revealing the function's behavior at a more detailed level. Understanding derivatives is key to deeper mathematical analysis and solving complex problems.