Problem 23 - Brief Applied Calculus

Question

Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(z)=\left(z^{4}+z^{2}+1\right)\left(z^{3}-z\right) $$

Step-by-Step Solution

Answer

The derivative is \( 7z^6 - 1 \).

Step 1: Identify the functions

In the given function, identify the two functions that are being multiplied. Let \( u(z) = z^4 + z^2 + 1 \) and \( v(z) = z^3 - z \). These are the two functions for which we will apply the Product Rule.

Step 2: Apply the Product Rule

The Product Rule states that the derivative of a product of two functions \( u(z) \) and \( v(z) \) is given by \( f'(z) = u'(z)v(z) + u(z)v'(z) \).

Step 3: Find the derivative of u(z)

To find \( u'(z) \), take the derivative of \( u(z) = z^4 + z^2 + 1 \). Using basic derivative rules, \( u'(z) = 4z^3 + 2z \).

Step 4: Find the derivative of v(z)

To find \( v'(z) \), take the derivative of \( v(z) = z^3 - z \). Using basic derivative rules, \( v'(z) = 3z^2 - 1 \).

Step 5: Substitute into the Product Rule

Substitute \( u'(z) = 4z^3 + 2z \), \( v(z) = z^3 - z \), \( u(z) = z^4 + z^2 + 1 \), and \( v'(z) = 3z^2 - 1 \) into the product rule equation to get:\[ f'(z) = (4z^3 + 2z)(z^3 - z) + (z^4 + z^2 + 1)(3z^2 - 1) \]

Step 6: Simplify the terms

Expand each term. - For \( (4z^3 + 2z)(z^3 - z) \): \[ 4z^3 \cdot z^3 + 4z^3 \cdot (-z) + 2z \cdot z^3 + 2z \cdot (-z) = 4z^6 - 4z^4 + 2z^4 - 2z^2 = 4z^6 - 2z^4 - 2z^2 \]- For \( (z^4 + z^2 + 1)(3z^2 - 1) \): \[ z^4 \cdot 3z^2 + z^4 \cdot (-1) + z^2 \cdot 3z^2 + z^2 \cdot (-1) + 1 \cdot 3z^2 + 1 \cdot (-1) = 3z^6 - z^4 + 3z^4 - z^2 + 3z^2 - 1 = 3z^6 + 2z^4 + 2z^2 - 1 \]

Step 7: Combine and simplify the expression

Combine the expanded expressions from Step 6:\[ 4z^6 - 2z^4 - 2z^2 + 3z^6 + 2z^4 + 2z^2 - 1 = 7z^6 - 1 \]This is the fully simplified expression for \( f'(z) \).

Key Concepts

Product Rule

In calculus, the Product Rule is a fundamental technique for finding the derivative of the product of two functions. If you have two functions, say \( u(z) \) and \( v(z) \), and you want to find the derivative of their product, you can use this rule. The Product Rule states that the derivative \( f'(z) \) of the product \( u(z) \cdot v(z) \) is given by:

• Find the derivative of the first function \( u'(z) \).
• Multiply \( u'(z) \) by the original second function \( v(z) \).
• Next, find the derivative of the second function \( v'(z) \).
• Multiply \( v'(z) \) by the original first function \( u(z) \).
• Add the two resulting products together.

This can be succinctly expressed as:\[f'(z) = u'(z)v(z) + u(z)v'(z)\] By following these steps, you can find the derivative of a product without having to multiply the functions outright before differentiating. This method helps save both time and effort, especially with complex expressions.

Derivative Simplification

Derivative simplification involves reducing complex expressions into a simpler and more manageable form. After applying the Product Rule, derivatives can often result in cumbersome expressions. Simplifying these expressions helps in both understanding and further manipulation. Here's how you can simplify derivatives:

• Distribute terms appropriately when multiplying expressions. This involves applying the distributive property, where you multiply each term in one expression by each term in the other.
• Combine like terms. This means grouping terms that have the same variables and powers together, and then summing or subtracting these groups as necessary.
• Cancel out terms where possible. Sometimes, due to simplification, certain terms may directly cancel out or reduce to zero.

By carefully expanding and combining these terms, what initially seems complex can often boil down to a much simpler expression. This process is crucial for making further calculations more straightforward and for transparent mathematical communication.

Polynomial Functions

Polynomial functions are a key type of function in calculus, and understanding them is essential for mastering derivatives. A polynomial is a function that is expressed as the sum of terms consisting of a variable raised to a nonnegative integer exponent, each multiplied by a coefficient. For example, polynomial functions take forms such as \( a_nz^n + a_{n-1}z^{n-1} + \ldots + a_1z + a_0 \), where \( a_n, a_{n-1}, \ldots, a_0 \) are coefficients.Polynomials have various degrees, defined by the highest power of the variable they contain. In the context of derivatives:

• Finding the derivative of a polynomial involves applying basic differentiation rules to each term separately. The derivative of \( z^n \) is \( nz^{n-1} \).
• Derivatives of polynomials are often simpler than derivatives of other kinds of functions since each term is dealt with individually and follows straightforward rules.
• Simplified derivatives of polynomials are also polynomials; however, they are one degree lower than the original, because differentiation reduces the power of each term by one.

Polynomial functions are among the simplest to differentiate, making them a perfect starting point for learning calculus derivatives.