Chapter 3: Problem 26
\(3-26\) Differentiate. $$f(x)=\frac{a x+b}{c x+d}$$
Short Answer
Expert verified
The derivative is \( f'(x) = \frac{ad - bc}{(cx + d)^2} \).
Step by step solution
01
Understand the problem
We need to find the derivative of the function \( f(x) = \frac{a x + b}{c x + d} \). This function is in the form of a quotient \( \frac{u}{v} \), where \( u = ax + b \) and \( v = cx + d \).
02
Identify the differentiation rule
Since \( f(x) \) is a quotient, we will use the quotient rule for differentiation, which states: If \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \).
03
Differentiate the numerator and denominator
First, compute the derivatives of the numerator and the denominator: \( u = ax + b \Rightarrow u' = a \) (as the derivative of a linear function is the coefficient) and \( v = cx + d \Rightarrow v' = c \).
04
Apply the quotient rule
Plug in the derivatives into the quotient rule: \[ f'(x) = \frac{(a)(cx + d) - (ax + b)(c)}{(cx + d)^2} \].
05
Simplify the expression
Expand and simplify the expression: \[ f'(x) = \frac{acx + ad - acx - bc}{(cx + d)^2} \]. The \( acx \) terms cancel out, yielding: \[ f'(x) = \frac{ad - bc}{(cx + d)^2} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a fundamental technique in calculus used to differentiate functions that are expressed as a fraction of two other functions. Imagine you have a function written as a quotient like \( f(x) = \frac{u}{v} \). Here, \( u \) and \( v \) are both differentiable functions of \( x \).
When we want to find the derivative of such a function, we use the quotient rule which states:
It is essential in calculus whenever you encounter a quotient of functions as it simplifies the process and ensures accuracy.
When we want to find the derivative of such a function, we use the quotient rule which states:
- First, take the derivative of the numerator \( u \) to get \( u' \).
- Next, take the derivative of the denominator \( v \) to get \( v' \).
- Then apply the formula for the derivative: \( f'(x) = \frac{u'v - uv'}{v^2} \).
It is essential in calculus whenever you encounter a quotient of functions as it simplifies the process and ensures accuracy.
Linear Function Differentiation
A linear function in its simplest form can be expressed as \( u(x) = ax + b \). This is a straight line graph where \( a \) is the slope and \( b \) is the y-intercept.
Differentiating a linear function is simple and straightforward:
In our exercise, both the numerator and the denominator of the function form linear functions \( u = ax + b \) and \( v = cx + d \). Thus, their derivatives are \( u' = a \) and \( v' = c \) respectively. With such functions, this simple linear differentiation proceeds seamlessly into more complex processes like applying the quotient rule.
Differentiating a linear function is simple and straightforward:
- Take the derivative of \( ax + b \), which is simply \( a \).
- This means the rate of change of the linear function is constant and equal to the slope \( a \).
In our exercise, both the numerator and the denominator of the function form linear functions \( u = ax + b \) and \( v = cx + d \). Thus, their derivatives are \( u' = a \) and \( v' = c \) respectively. With such functions, this simple linear differentiation proceeds seamlessly into more complex processes like applying the quotient rule.
Derivative of a Quotient
To find the derivative of a quotient of two functions, like \( f(x) = \frac{u}{v} \), you combine both the quotient rule and the process of linear differentiation.
In the exercise, the function to differentiate is \( f(x) = \frac{ax+b}{cx+d} \). This makes \( u = ax + b \) and \( v = cx + d \).
First, differentiate each component:
\[ f'(x) = \frac{(a)(cx + d) - (ax + b)(c)}{(cx + d)^2} \] Expand and simplify this to:
\[ f'(x) = \frac{ad - bc}{(cx + d)^2} \] This expression shows the rate of change of the quotient function.
Simplifying correctly is crucial to understanding internal cancellations like \( acx - acx \), which help to streamline the derivative.
In the exercise, the function to differentiate is \( f(x) = \frac{ax+b}{cx+d} \). This makes \( u = ax + b \) and \( v = cx + d \).
First, differentiate each component:
- The derivative of the numerator is \( u' = a \).
- The derivative of the denominator is \( v' = c \).
\[ f'(x) = \frac{(a)(cx + d) - (ax + b)(c)}{(cx + d)^2} \] Expand and simplify this to:
\[ f'(x) = \frac{ad - bc}{(cx + d)^2} \] This expression shows the rate of change of the quotient function.
Simplifying correctly is crucial to understanding internal cancellations like \( acx - acx \), which help to streamline the derivative.