Problem 28 - Calculus for Biology and Medicine

Question

Apply the product rule for the product of three functions to find the derivative of \(y=f(x)\). \(f(x)=(2 x+1)\left(4-x^{2}\right)\left(1+x^{2}\right)\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -10x^4 - 4x^3 + 18x^2 + 6x + 8 \).

Step 1: Identify the Product

We begin by recognizing that the function is a product of three functions: \(u(x) = 2x + 1\), \(v(x) = 4 - x^2\), and \(w(x) = 1 + x^2\). These are the functions we need to differentiate using the product rule for three functions.

Step 2: Recall the Product Rule for Three Functions

The derivative of a product of three functions \( f(x) = u(x) \,v(x) \,w(x)\) is given by the formula: \[ f'(x) = u'(x) \, v(x) \, w(x) + u(x) \, v'(x) \, w(x) + u(x) \, v(x) \, w'(x) \] This rule involves taking the derivative of each function one at a time, and keeping the other two functions as they are, then adding these results together.

Step 3: Find Derivatives of Each Function

Now, we find the derivative of each individual function. Start with: - \( u(x) = 2x + 1 \), so \( u'(x) = 2 \) - \( v(x) = 4 - x^2 \), so \( v'(x) = -2x \) - \( w(x) = 1 + x^2 \), so \( w'(x) = 2x \)

Step 4: Apply the Product Rule

Using the derivatives calculated in the previous step, substitute them into the product rule for three functions: \[ f'(x) = u'(x) \, v(x) \, w(x) + u(x) \, v'(x) \, w(x) + u(x) \, v(x) \, w'(x) \] Substituting we get: \[ f'(x) = (2) (4 - x^2) (1 + x^2) + (2x + 1) (-2x) (1 + x^2) + (2x + 1) (4 - x^2) (2x) \]

Step 5: Simplify Each Term

We simplify each term: 1. \( 2(4-x^2)(1+x^2) = 2(4 - x^4 + 3x^2) = 8 + 6x^2 - 2x^4 \) 2. \( (2x+1)(-2x)(1 + x^2) = (-4x^2 - 2x)(1 + x^2) = -4x^2 - 4x^4 - 2x - 2x^3 \) 3. \( (2x+1)(4-x^2)(2x) = (2x+1)(8x - 2x^3) = 16x^2 - 4x^4 + 8x - 2x^3 \)

Step 6: Combine and Simplify Terms

Combine all terms from each part: Combining, we add all the results together: \[ 8 + 6x^2 - 2x^4 - 4x^2 - 4x^4 - 2x - 2x^3 + 16x^2 - 4x^4 + 8x - 2x^3 \] Combine like terms: - Constant terms: \( 8 \) - \( x \) terms: \( 8x - 2x = 6x \) - \( x^2 \) terms: \( 6x^2 - 4x^2 + 16x^2 = 18x^2 \) - \( x^3 \) terms: \( -2x^3 - 2x^3 = -4x^3 \) - \( x^4 \) terms: \( -2x^4 - 4x^4 - 4x^4 = -10x^4 \) Therefore, \[ f'(x) = -10x^4 - 4x^3 + 18x^2 + 6x + 8 \]

Key Concepts

Derivative: The Concept of Change

In calculus, the derivative is a fundamental concept that measures how a function changes as its input changes. Think of the derivative as the "rate of change" or "slope" of the function. It tells us how fast the function's value is changing at any given point. This is crucial for understanding not just mathematical concepts, but also real-world phenomena like speed, growth rates, and other changes over time.

For a single-variable function like \(f(x)\), the derivative \(f'(x)\) represents the slope of the tangent line to the function's graph at any point \(x\). If \(f(x)\) represents a curve on a graph, \(f'(x)\) is the slope of that curve at \(x\).

In our given exercise, we are tasked with finding the derivative of a more complex function, which involves applying specific rules to handle the product of multiple functions.

Three Functions: Handling the Complexity

When dealing with functions that are products of multiple terms, such as the three functions in our exercise, it's necessary to apply specific rules to find their derivatives. In this case, we use the product rule, extended to handle three functions. The product rule for two functions is already a powerful tool, but when adding an extra function into the mix, the rule expands accordingly.

We begin by identifying our functions as follows:

• \(u(x) = 2x + 1\)
• \(v(x) = 4 - x^2\)
• \(w(x) = 1 + x^2\)

The product rule for three functions suggests that the derivative of a product \(f(x) = u(x) \, v(x) \, w(x)\) is given by:
\[ f'(x) = u'(x) \, v(x) \, w(x) + u(x) \, v'(x) \, w(x) + u(x) \, v(x) \, w'(x) \]

This formula involves taking the derivative of one function at a time (leaving the other two as they are), then summing up these derivatives, which allows us to handle the complexity of multiple function products effectively.

Simplifying Terms: Bringing it All Together

Once we've applied the product rule and found the derivatives of the individual terms involved, the next essential step is simplifying. Simplifying terms is crucial to make the final derivative expression more interpretable and easier to work with.

After finding the individual derivatives and applying the product rule, we substitute the derived values into the expanded formula:
\[ f'(x) = (2) (4 - x^2) (1 + x^2) + (2x + 1) (-2x) (1 + x^2) + (2x + 1) (4 - x^2) (2x) \]

We simplify each term individually:

• The first term simplifies to \(8 + 6x^2 - 2x^4\).
• The second becomes \(-4x^2 - 4x^4 - 2x - 2x^3\).
• The third term simplifies to \(16x^2 - 4x^4 + 8x - 2x^3\).

Finally, we combine all these simplified terms, looking for like terms, which gives us the final, simplified derivative:
\[ f'(x) = -10x^4 - 4x^3 + 18x^2 + 6x + 8 \]

This final expression tells us the complete rate of change of the original function with respect to \(x\), gathered and simplified for clarity.