Problem 82 - Calculus for Biology and Medicine

Question

Suppose that \(f(2)=-4\), and \(f^{\prime}(2)=1\). Let \(y=1 / f(x)\); find \(\frac{d y}{d x}\) when \(x=2\).

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) at \( x = 2 \) is \(-\frac{1}{16}\).

Step 1: Understand the Problem

We are given a function \( f(x) \) such that \( f(2) = -4 \) and its derivative \( f'(2) = 1 \). We need to find the derivative of \( y = \frac{1}{f(x)} \) at \( x = 2 \).

Step 2: Recall the Derivative Formula for a Function Involving a Quotient

The derivative of \( y = \frac{1}{f(x)} \) can be found using the quotient rule. This is given by \( \frac{d}{dx}\left(\frac{1}{u}\right) = -\frac{u'}{u^2} \), where \( u = f(x) \) and \( u' = f'(x) \).

Step 3: Apply the Derivative Formula

Substitute \( u = f(x) \) and \( u' = f'(x) \) into the formula from Step 2: \[ \frac{dy}{dx} = -\frac{f'(x)}{(f(x))^2} \].

Step 4: Substitute the Given Values at \( x = 2 \)

Use the values \( f(2) = -4 \) and \( f'(2) = 1 \) in the derivative formula: \[ \frac{dy}{dx} = -\frac{1}{(-4)^2} = -\frac{1}{16} \].

Step 5: Simplify the Expression

Calculate the derivative expression: \[ \frac{dy}{dx} = -\frac{1}{16} \]. This is the rate of change of \( y = \frac{1}{f(x)} \) at \( x = 2 \).

Key Concepts

Quotient Rule in Derivatives

When you work with derivatives involving division of two functions, the quotient rule is essential. It helps us find the derivative of functions in the form of a quotient, specifically where one function is divided by another. Here is how it works.

The quotient rule states that if you have functions \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u(x)}{v(x)} \) is:

• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

For our problem, we used a specific form of the quotient rule because \( y = \frac{1}{f(x)} \) can be seen as \( y = \frac{1}{u} \). The derivative formula simplifies to \( -\frac{u'}{u^2} \) for this specific case.

Understanding this rule can significantly ease the task of finding how one function changes relative to another, especially in more complex calculus problems.

Understanding Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. It's like finding the slope of a line tangent to a function's curve at a specific location. This concept is a fundamental building block to understanding how things change.

Mathematically, the derivative of a function \( f(x) \), at a particular point \( x \/ \), is given by \( f'(x) \). It represents the function's instantaneous rate of change or the slope of the tangent line to the graph of \( f \) at \( x \/ \).

In our exercise, we had \( f'(2) = 1 \), which tells us the rate of change of \( f(x) \/ \) is 1 when \( x = 2 \/ \). To find the derivative of \( y = \frac{1}{f(x)} \), we use this concept along with the quotient rule to discover how \( y \/ \) changes at \( x = 2 \/ \). Understanding derivatives is crucial for solving real-world problems where rates of change affect decisions.

Evaluating Functions

Function evaluation is the process of finding the output of a function for a specific input. It is a fundamental arithmetic operation required to understand the behavior of functions at particular points.

To evaluate a function, we simply substitute the input value into the function. For instance, evaluating \( f(x) \) at \( x = 2 \) gives \( f(2) = -4 \). This procedure provides us the specific value of the function at that input, which is essential for further calculations.

In the exercise, knowing \( f(2) \) and \( f'(2) \) directly influenced the solution process. Substituting these into derivative formulas allowed us to determine how \( y \) changes at \( x = 2 \). Function evaluation lets us make precise calculations and is a key step in solving many mathematical and real-world problems.