Problem 26 - Calculus for Biology and Medicine

Question

Apply the product rule repeatedly to find the derivative of \(y=f(x) .\) $$ f(x)=(x-3)(2-3 x)(5-x) $$

Step-by-Step Solution

Answer

The derivative of \( f(x) \) is calculated by applying and simplifying the product rule: \( f'(x) = (11-6x)(5-x) - (x-3)(2-3x) \).

Step 1: Understand the Product Rule

The product rule in calculus is used to find the derivative of a product of two functions. If we have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product is given by \( (uv)' = u'v + uv' \). When we have more than two functions, we apply the product rule repeatedly.

Step 2: Assign Functions to Product

Given the function \( f(x) = (x-3)(2-3x)(5-x) \), assign: \( u(x) = (x-3) \), \( v(x) = (2-3x) \), and \( w(x) = (5-x) \). We will treat this as a product of three functions, \( (uv) \cdot w \).

Step 3: Differentiate the First Product

First, find the derivative of \( u(x) \cdot v(x) \) using the product rule. Let \( u(x) = (x-3) \) and \( v(x) = (2-3x) \). Then, \( u'(x) = 1 \) and \( v'(x) = -3 \). The derivative is: \( (uv)' = u'v + uv' = 1ig(2-3x\big) + (x-3)(-3) \).

Step 4: Simplify the First Product Derivative

Simplify the expression from Step 3: \( (uv)' = (2-3x) + (-3x+9) \), which simplifies to: \( (uv)' = 11 - 6x \).

Step 5: Apply Product Rule to Full Product

Now apply the product rule to the entire original function as \( (uv)w \):\( f'(x) = (uv)'w + (uv)w' \).

Step 6: Calculate Derivative of the Full Product

We have \( (uv)' = 11 - 6x \) and from Step 2, \( w'(x) = -1 \). Substitute these derivatives plus \((uv) = (x-3)(2-3x) \) and \( w(x) = (5-x) \) into the expression:\( f'(x) = (11-6x)(5-x) + (x-3)(2-3x)(-1) \).

Step 7: Simplify the Complete Expression

Expand and simplify: \( f'(x) = (11-6x)(5-x) - (x-3)(2-3x) \) After multiplying and combining like terms, you will arrive at the final simplified derivative.

Key Concepts

Understanding Derivatives

In calculus, the derivative is a fundamental concept that measures how a function changes as its input changes. If you think of a graph of a function, the derivative at a certain point gives you the slope of the tangent line to the graph at that point. This is like finding out how steep a hill or road is. For example, if a function represents the distance a car travels over time, the derivative at any point would give you the car's speed at that exact moment.

• The steeper the slope, the greater the rate of change.
• If the slope is flat (derivative is zero), the function's value is not changing at that point.
• If the derivative is negative, it indicates a decrease in the function's value.

Understanding derivatives is crucial because they provide insights into the behavior of functions, such as where they increase, decrease, or have any kind of inflection (change in curvature). In the context of the product rule, applying derivatives allows us to understand the rate of change for more complex expressions, like products of multiple functions.

Exploring Calculus

Calculus is a branch of mathematics focused on change and motion. It allows us to model and study real-world processes that exhibit continuous change. Introduced in the 17th century, calculus has dramatically expanded our ability to solve problems across various scientific fields.
There are two main branches of calculus:

• Differential Calculus: Concerns itself with the concept of a derivative and how functions change.
• Integral Calculus: Deals with the accumulation of quantities and the areas under curves.

Differential calculus involves notions like limits and instantaneous rates of change, which we see in the use of derivatives. The product rule, an important derivative formula, showcases how calculus breaks down complex problems into simpler, more manageable pieces. By understanding how the derivative of a product is calculated, students can handle various functions that model multiplicative interactions.

Differentiation of Products Using the Product Rule

When differentiating products of two or more functions, the product rule is an essential tool. It tells us how to find the derivative of a product of functions by breaking it down into simpler parts. If you're dealing with two functions, say \( u(x) \) and \( v(x) \), the product rule formula is: \( (uv)' = u'v + uv' \).
With more than two functions, like in the exercise given with \( (x-3)(2-3x)(5-x) \), you apply the product rule repeatedly.
Here's how it works:

• Find the derivative of the first pair of functions. Let's denote these as \( u(x) = (x-3) \) and \( v(x) = (2-3x) \). Calculate \( (uv)' \) by combining their derivatives.
• Simplify \( (uv)' \).
• Treat \( (uv) \) as a single function and apply the product rule to \( (uv) \) and the third function \( w(x) = (5-x) \).
• Combine everything into the final expression for the derivative \( f'(x) \).

This method allows us to extract the derivative of complex, multi-term functions using systematic breakdown and simplification, making the challenging world of calculus more approachable.