Question

Finding a Derivative In Exercises \(25-38\) , find the derivative of the algebraic function. $$ f(x)=\left(2 x^{3}+5 x\right)(x-3)(x+2) $$

Short answer

The derivative of the function \(f(x)= (2x^3 + 5x)(x-3)(x+2)\) is \(f'(x) = 12x^4 + 6x^3 - 30x^2 - 66x + 15\).

Step-by-step solution

Identify the Functions and Their Derivatives

Identify \(u = 2x^3 + 5x\), \(v = (x-3)(x+2)\), then find \(u'\) and \(v'\). Applying the power rule we find that \(u' = 6x^2 + 5\). For \(v = (x-3)(x+2)\), apply the power rule and the simple rule that states the derivative of a constant is zero: \(v' = (1)(x+2) + (x-3)(1) = x+2 + x-3 = 2x - 1\)

Apply the Product Rule

The product rule says that the derivative of the product of two functions is \(u'v + uv'\). Using the results from Step 1, \(u'v = (6x^2 + 5)(x-3)(x+2)\) and \(uv' = (2x^3 +5x)(2x-1)\). This means that \(f'(x) = (6x^2 + 5)(x-3)(x+2) + (2x^3 +5x)(2x-1)\). Then distribute each term and simplify.

Distribute and Simplify

Now, expand and simplify \(f'(x)\). After distributing and combining like terms, we get: \(f'(x) = 12x^4 + 6x^3 - 30x^2 - 66x + 15\).

Key concepts

Product Rule

The product rule is a fundamental tool in calculus used to find derivatives of equations that are products of two or more functions. Let's say you have two functions, \( u \) and \( v \). The product rule states that the derivative of the product \( uv \) is given by \( u'v + uv' \). Here's a simple breakdown of this process:

• First, differentiate \( u \) to find \( u' \), and differentiate \( v \) to find \( v' \).
• Then, multiply \( u' \) by \( v \), and \( u \) by \( v' \).
• Finally, sum these two products to get the derivative of \( uv \).

Applying the product rule allows one to solve complex derivatives by breaking them down into simpler parts.

Power Rule

The power rule is a quick way to find the derivative of a function in the form of \( x^n \). If \( f(x) = x^n \), the power rule tells us that the derivative \( f'(x) = nx^{n-1} \). Let’s break down the steps:

• Identify the exponent \( n \) in \( x^n \).
• Multiply \( n \) by the coefficient of \( x \) (if there is one).
• Reduce the exponent by one.

For example, if you want to differentiate \( 2x^3 \), applying the power rule, you multiply 2 by 3 to get 6, then reduce the power of x from 3 to 2, resulting in \( 6x^2 \). This rule is particularly useful in handling polynomial functions.

Algebraic Functions

Algebraic functions are combinations of polynomials that can include operations such as addition, subtraction, multiplication, and division. These functions can be quite simple or complex and are characterized by terms that involve variables raised to various powers. In the context of derivatives, these functions often require applying rules such as the product and power rules to simplify differentiation.

• Polynomials like \( 2x^3 + 5x \) are straightforward algebraic functions.
• Expressions that include products or ratios of polynomials, like \( (x-3)(x+2) \), need a bit more effort to differentiate properly.

A solid understanding of algebra is crucial for working with these functions in calculus.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative provides critical insights into how a function changes for small changes in its input value, which is essential in fields such as physics, economics, and engineering.

• The main goal is to find the rate of change or slope of a function at a given point.
• Utilize various rules like the product, power, and chain rules to break down complex functions.
• Simplification is often required after applying these rules, demanding careful algebraic manipulation.

In essence, differentiation is a key calculation in calculus that helps describe real-world scenarios by understanding changes.