Problem 35 - Calculus for Biology and Medicine

Question

Suppose that \(f(2)=-4, g(2)=3, f^{\prime}(2)=1\), and \(g^{\prime}(2)=-2\). Find $$ (f g)^{\prime}(2) $$

Step-by-Step Solution

Answer

The derivative \((fg)'(2)\) is 11.

Step 1: Identify the Rule

To find the derivative of the product of two functions, we will use the product rule. The product rule states that if we have two functions \(f(x)\) and \(g(x)\), the derivative of their product is given by \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\).

Step 2: Apply the Product Rule

Using the product rule, we have \((fg)'(x) = f'(x)g(x) + f(x)g'(x)\). Therefore, at \(x=2\), \((fg)'(2) = f'(2)g(2) + f(2)g'(2)\).

Step 3: Substitute the Given Values

Substitute the provided values into the equation: \(f'(2) = 1\), \(g(2) = 3\), \(f(2) = -4\), and \(g'(2) = -2\). This gives us \((fg)'(2) = 1 \times 3 + (-4) \times (-2)\).

Step 4: Simplify the Expression

Calculate the expression step by step: \(1 \times 3 = 3\) and \((-4) \times (-2) = 8\). Thus, \((fg)'(2) = 3 + 8\).

Step 5: Calculate the Final Result

Add the two results together to get \((fg)'(2) = 3 + 8 = 11\).

Key Concepts

Understanding the Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a function changes at any given point. It tells us how much a function's output changes with respect to a change in input. Think of it as a way of measuring how fast something is moving or changing. For a given function, the derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \). It's like the slope of a line at a specific point on a curve.

• For a straight line, the derivative is constant. This makes sense because the slope doesn’t change.
• For a parabola, like \( y = x^2 \), the derivative increases linearly with \( x \).

When working with derivatives, we can understand both linear and non-linear changes and predict future behavior of mathematical models.

Exploring Functions

Functions are the backbone of calculus and describe relationships between variables. In the context of derivatives, functions allow us to define expressions and analyze their behavior over different intervals. A function assigns one output for every input, often written as \( f(x) \). For example, if \( f(x) = x^2 \), then \( f(3) = 9 \).Functions are incredibly powerful because they give us a way to model real-world scenarios. Here's what makes them so essential:

• They represent complex systems, from physics to finance.
• They have different types, such as linear, quadratic, and exponential, each with unique characteristics.
• Understanding their behavior in various contexts, like finding maximum and minimum points, is crucial.

When you take the derivative of a function, you gain insight into how the system it models behaves.

Differentiation Rules: The Product Rule

Differentiation is the process of finding a derivative, and there are several rules that simplify this process. One key rule is the product rule, which comes into play when dealing with the derivative of the product of two functions. The product rule states:\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \]Let's break it down:

• \( f'(x) \) represents the derivative of the first function.
• \( g(x) \) is the second function as is.
• \( f(x) \) is the first function untouched.
• \( g'(x) \) is the derivative of the second function.

When you apply this rule, you effectively calculate the rate of change for each function in relation to the other. The product rule is beneficial for solving problems involving products of functions, providing a clear pathway to the derivative where otherwise, the operation would be complex.