Question

Consider the following functions and express the relationship between a small change in \(x\) and the corresponding change in \(y\) in the form \(d y=f^{\prime}(x) d x\). \(f(x)=(4+x) /(4-x)\)

Short answer

Question: Given the function \(f(x) = (4+x)/(4-x)\), find the relationship between a small change in \(x\) and the corresponding change in \(y\). Answer: For the given function, the relationship between a small change in \(x\) and the corresponding change in \(y\) is given by \(dy = \frac{8}{(4-x)^2} dx\).

Step-by-step solution

Find the derivative of the given function

First, we need to find the derivative of \(f(x) = (4+x)/(4-x)\). To do this, we can use the quotient rule, which states that if we have a function in the form of \(h(x) = \frac{g(x)}{k(x)}\), then the derivative \(h'(x)\) can be calculated as: \(h'(x) = \frac{g'(x)k(x) - g(x)k'(x)}{[k(x)]^2}\) So for our function \(f(x) = (4+x)/(4-x)\), we have: \(g(x) = 4+x\) \(g'(x) = 1\) \(k(x) = 4-x\) \(k'(x) = -1\) Now we can apply the quotient rule: \(f'(x) = \frac{g'(x)k(x)-g(x)k'(x)}{[k(x)]^2} = \frac{1(4-x)-(4+x)(-1)}{(4-x)^2}\) Simplify the expression, we get: \(f'(x) = \frac{4 - x + 4 + x}{(4-x)^2} = \frac{8}{(4-x)^2}\)

Express the change in \(y\) in terms of the change in \(x\)

Now that we have found the derivative, \(f'(x) = \frac{8}{(4-x)^2}\), we can express the small change in \(y\) as a function of the small change in \(x\), represented as \(dx\). This can be done using the following relationship: \(dy = f'(x) dx\) So for our function, we have: \(dy = \frac{8}{(4-x)^2} dx\) And this is the required relationship between a small change in \(x\) and the corresponding change in \(y\).

Key concepts

Derivative

A derivative represents the rate at which a function changes at any point. Think of it as a way to determine how steep a curve is at a specific point. Mathematically, it is expressed as the limit of the difference quotient as the interval approaches zero. This means, it's about how much the function's output values change in response to a change in its input values.

In our exercise, to find the derivative of the function \( f(x) = \frac{4+x}{4-x} \), we use the quotient rule. This is necessary because our function is a ratio of two functions. The concept of the derivative is central to understanding how to express one variable's rate of change with respect to another, essentially connecting small changes in \(x\) to changes in \(y\).

The Quotient Rule comes into play when calculating derivatives of functions that are quotients, and its formula is:

• \( h'(x) = \frac{g'(x)k(x) - g(x)k'(x)}{(k(x))^2} \)

In this rule, we have two functions, \(g(x)\) for the numerator and \(k(x)\) for the denominator. Using this, we can break down the complex problem into parts that are easier to solve.

Differentiation

Differentiation is the process of finding a derivative. It is a key element in calculus and helps us determine the slope of a function at any given point. This means that through differentiation, one can find how rapidly or slowly something changes.

During differentiation, you apply rules like the quotient rule or product rule. Here in our exercise, by applying the quotient rule correctly, we derived

• \( f'(x) = \frac{8}{(4-x)^2} \)

This means that the function's rate of change at any point \(x\) in the domain can be expressed as \(8/(4-x)^2\). It can also tell us other things like where a function is increasing or decreasing.

Thus, through differentiation, we can provide insights about direct relationships between variables, just like encoding the relationship of changes in \(x\) which help us understand changes in \(y\).

Chain Rule

The Chain Rule allows us to differentiate composite functions or functions within functions. When you see a function that is nested within another, like \((4+x)/(4-x)\) could possibly transform into more complex nested forms, the chain rule helps us differentiate them effectively and efficiently.

Although the simple form of the given function doesn’t directly require the chain rule, complex functions that have such a structure benefit greatly from this fundamental rule. For a function \( u(v(x)) \), the derivative is determined as \( u'(v(x)) \times v'(x) \). This rule simplifies taking derivatives of complicated expressions by breaking them into easier parts.

In the function from the exercise, despite mainly using the quotient rule, being aware of the chain rule enables us to handle similar problems when functions become nested, especially useful in more advanced differentiation tasks in calculus.