Problem 42 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable at \(x .\) Find an expression for the derivative of \(y .\) \(y=[f(x)-3] g(x)\)

Step-by-Step Solution

Answer

The derivative of \(y\) is \(y' = f'(x)g(x) + f(x)g'(x) - 3g'(x)\).

Step 1: Identify the Rule to Use

To find the derivative of the product of two differentiable functions, we use the product rule. If we have a function in the form of \(y = u(x) \cdot v(x)\), then the derivative \(y'\) is given by \(y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)\).

Step 2: Define the Parts of the Product

In our given function \(y = [f(x) - 3]g(x)\), let's identify \(u(x)\) and \(v(x)\). We have \(u(x) = f(x) - 3\) and \(v(x) = g(x)\).

Step 3: Find the Derivatives of Each Part

We need to find the derivatives of \(u(x)\) and \(v(x)\). - For \(u(x) = f(x) - 3\), the derivative is \(u'(x) = f'(x)\) because the derivative of a constant is zero.- For \(v(x) = g(x)\), the derivative is \(v'(x) = g'(x)\).

Step 4: Apply the Product Rule

Now, substitute \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) into the product rule formula:\[y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) = f'(x) \cdot g(x) + (f(x) - 3) \cdot g'(x)\].

Step 5: Simplify the Expression

Combine the terms of the derivative:\[y' = f'(x) \cdot g(x) + f(x) \cdot g'(x) - 3 \cdot g'(x)\].This is the simplified form of the expression for the derivative of \(y\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves calculating the derivative of a function. It helps us understand how a function changes, or in simpler terms, the rate at which one quantity changes with respect to another. To differentiate a function, we essentially follow set rules to find this rate of change. The process is particularly useful in a variety of fields, ranging from physics to economics, where it helps to predict trends and demands.

Here are some key points to remember about differentiation:

• The derivative of a constant is always zero, since a constant doesn’t change.
• Differentiation rules let us find the derivative of more complex expressions, like sums, differences, and products of functions.

In the given exercise, we see differentiation in action, as the product rule is used to differentiate the product of two functions. This rule lets us find the derivative of a product by considering the derivatives of individual functions involved.

Derivative

A derivative represents the rate of change of a function as its input changes. It’s commonly symbolized as \(f'(x)\), \(y'\), or \(\frac{dy}{dx}\). Think of the derivative as a mathematical way of finding the slope of the tangent line to a curve at any given point.

Here’s what you need to know about derivatives:

• A derivative can tell us about the local behavior of functions, indicating whether it's increasing or decreasing at any given point.
• Different rules apply when finding derivatives of different types of functions. For example, the power rule, product rule, and chain rule simplify the process significantly.

In the exercise above, the derivative \(y'\) is determined by identifying individual parts of the function \(y = [f(x) - 3] \, g(x)\) and applying the product rule. This rule highlights how derivatives come together to express changes in compound functions.

Differentiable Functions

Differentiable functions are those that have a derivative at every point in their domain. Simply put, they are smooth and connected functions that do not have sharp corners or discontinuities. This smoothness allows us to apply calculus-based techniques like differentiation.

Key things to remember about differentiable functions include:

• If a function is differentiable, it is also continuous. However, continuity alone doesn’t imply differentiability.
• Common operations like addition, subtraction, and multiplication usually maintain differentiability between functions.

In our exercise, both \(f(x)\) and \(g(x)\) are differentiable, which ensures their product is also differentiable. Applying the product rule rests on the assumption that these functions are differentiable, thus guaranteeing that their derivative expressions are valid throughout their domains.