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

Step 1: Recognize the Product Rule

To find the derivative of the product of two functions, we need to use the product rule. The product rule states that if you have two functions, say \(u(x)\) and \(v(x)\), then the derivative of their product \(u(x) v(x)\) is \(u'(x) v(x) + u(x) v'(x)\).

Step 2: Identify Functions and Their Derivatives

Let \(u(x) = f(x) - 3\) and \(v(x) = g(x)\). We need to find \(u'(x)\) and \(v'(x)\). Since \(u(x) = f(x) - 3\), its derivative \(u'(x)\) is simply \(f'(x)\) because the derivative of a constant is zero. The derivative of \(v(x) = g(x)\) is \(v'(x) = g'(x)\).

Step 3: Apply the Product Rule

Using the product rule from Step 1, substitute the expressions from Step 2: The derivative of \(y = (f(x) - 3) g(x)\) is \(y' = (f(x) - 3)' g(x) + (f(x) - 3) g'(x)\).

Step 4: Substitute and Simplify

Substitute \(u'(x)\) and \(v'(x)\) into the product rule formula: \[ y' = f'(x) g(x) + (f(x) - 3) g'(x) \]This is the expression for the derivative of \(y\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes. It allows us to determine the slope of a curve at any given point, effectively telling us how steep the curve is. Think of differentiation as a tool to "zoom in" on any part of a curve to see its behavior in detail.

In the context of our exercise, the differentiation process is applied to a product of two functions, using the product rule. This is crucial because when functions are multiplied together, their derivatives can't simply be multiplied. Instead, we must use the product rule, which helps in breaking down the process incrementally.

• The derivative gives us significant insights into the function - identifying turning points, moments when the function is increasing, or when it's decreasing.
• In real-world applications, differentiation is used extensively in fields as diverse as physics, economics, and biology to model change and variation.

Derivative

A derivative represents the rate of change of a function with respect to a variable. In mathematical terms, for a function \( f(x) \), its derivative \( f'(x) \) tells us how \( f(x) \) changes as \( x \) changes. It can be understood as the slope of the tangent line to the curve at any point. This slope gives the best linear approximation to the curve at that point.

For our specific problem, we calculated the derivative of the product \( y = (f(x) - 3)g(x) \). Here:

• Derivative of \( (f(x) - 3) \) simplifies to \( f'(x) \), because differentiating a constant (like \(-3\)) results in zero.
• Derivative of \( g(x) \) is simply \( g'(x) \).

The product rule then combines these: \[ y' = f'(x)g(x) + (f(x) - 3)g'(x) \]
This expression tells us how the combined function \( y \) changes with \( x \). It’s a crucial piece of understanding dynamics in any situation where functions interact.

Mathematics

Mathematics offers structured ways of thinking about problems and allows us to express ideas in precise language. When it comes to calculus, it provides tools like differentiation to handle continuous change.

In our exercise, mathematics guided us through the use of the product rule to tackle a complex function. By applying mathematical principles:

• We can translate real-world problems into mathematical language, enabling solutions and predictions.
• Understanding concepts like the product rule is vital as it helps in simplifying and solving seemingly complex interactions between entities.

The beauty of mathematics lies in its applicability across different domains, enabling progress in science and technology. Through its logical structure, it gives us frameworks to dissect, comprehend, and innovate.