Problem 7 - Calculus for Biology and Medicine

Question

Use the product rule to find the derivative with respect to the independent variable. \(f(x)=\frac{1}{5}\left(x^{2}-1\right)\left(x^{2}+1\right)\)

Step-by-Step Solution

Answer

The derivative is \(f'(x) = \frac{4}{5}x^3\).

Step 1: Identify Functions in the Product

The given function is \(f(x)=\frac{1}{5}(x^2-1)(x^2+1)\). When applying the product rule, identify the two functions being multiplied: \(u(x) = x^2-1\) and \(v(x) = x^2+1\).

Step 2: Differentiate Each Function Separately

Differentiate \(u(x) = x^2-1\). The derivative is \(u'(x) = 2x\).Differentiate \(v(x) = x^2+1\). The derivative is \(v'(x) = 2x\).

Step 3: Apply the Product Rule

The product rule states that if \(f(x) = u(x)v(x)\), then \(f'(x) = u'(x)v(x) + u(x)v'(x)\).Apply this to the two functions:\[f'(x) = 2x(x^2+1) + (x^2-1)(2x)\].

Step 4: Expand and Simplify the Expression

First, expand each term:- \(2x(x^2+1) = 2x^3 + 2x\)- \((x^2-1)(2x) = 2x^3 - 2x\)Combine the terms:\[f'(x) = 2x^3 + 2x + 2x^3 - 2x\].

Step 5: Combine Like Terms

Combine the like terms to simplify the expression:\[f'(x) = 4x^3\].

Step 6: Include the Constant Factor

Remember to include the constant factor \(\frac{1}{5}\) from the original function:Thus, \[f'(x) = \frac{1}{5} \times 4x^3 = \frac{4}{5}x^3\].

Key Concepts

Discovering Derivatives

In calculus, derivatives are a fundamental concept that help us understand how a function behaves. The derivative of a function explains how the output value changes as the input value changes. It's like asking, "How quickly are we moving if we're traveling on a path represented by the function?"

In the context of our original exercise, the function given is a product of two simple functions:

• \( u(x) = x^2-1 \)
• \( v(x) = x^2+1 \)

To find the derivative, we need to look at how both these functions change. This is where the product rule comes in handy! The product rule allows us to find the derivative for functions that are multiplied together by considering each function separately first.

Understanding Functions

Functions are like machines. You input a number, and they give you a specific output. In the realm of calculus, understanding how functions work is key.

Every function consists of an independent variable and a dependent variable. In our case, \( x \) is the independent variable, determining the value of our function \( f(x) = \frac{1}{5}(x^2-1)(x^2+1) \). Here, our function combines two simpler parts, \( u(x) \) and \( v(x) \), into a single product.

• The first part, \( u(x) = x^2 - 1 \), decreases by 1 for every squared value of \( x \).
• The second part, \( v(x) = x^2 + 1 \), does the opposite, slightly increasing the squared value of \( x \).

Recognizing these components enables us to apply rules of calculus, like the product rule, to analyze the overall function's behavior.

The Power of Calculus

Calculus is the study of how things change and is often divided into two branches: differentiation and integration. Differentiation, which involves finding derivatives, allows us to understand the rate of change of functions.

In our exercise, calculus empowers us to apply the product rule, which is vital when dealing with the product of two or more functions. To use the product rule, remember:

• Differentiate each function individually.
• Apply the derivative of the first function times the second function plus the first function times the derivative of the second function.

Finally, simplify to get a neat expression. In our original problem, by using calculus, specifically the product rule, we find the derivative \( f'(x) = \frac{4}{5}x^3 \), which precisely tells us the rate at which our function \( f(x) \) changes as \( x \) changes.