Chapter 4: Problem 24
Find the derivatives of the functions. For extra practice, and to check your answers, do some of these in more than one way if possible. $$\frac{1}{1+1 / x}$$
Short Answer
Expert verified
The derivative is \( \frac{1}{(x+1)^2} \).
Step by step solution
01
Rewrite the Function
Rewrite the given function to a form that is easier to differentiate. We have \( y = \frac{1}{1 + \frac{1}{x}} \). This can be rewritten by getting a common denominator: \( y = \frac{x}{x + 1} \).
02
Differentiate using the Quotient Rule
Apply the quotient rule, which states that \( \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u = x \) and \( v = x+1 \). Find \( u' = 1 \) and \( v' = 1 \).
03
Apply Quotient Rule
Substitute \( u \), \( v \), \( u' \), and \( v' \) in the quotient rule formula: \[ y' = \frac{(x+1) \cdot 1 - x \cdot 1}{(x+1)^2} = \frac{x+1-x}{(x+1)^2} = \frac{1}{(x+1)^2} \].
04
Simplify the Derivative
The differentiated expression is \( \frac{1}{(x+1)^2} \). There's no further simplification needed as this is already in its simplest form.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
Differentiating functions that involve division requires a special technique called the quotient rule. Here’s where it comes into play: when you have a function of the form \( \frac{u}{v} \). The quotient rule helps us find the derivative of this function. The formula is
To illustrate, for the function \( y = \frac{x}{x + 1} \):
- \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)
To illustrate, for the function \( y = \frac{x}{x + 1} \):
- Set \(u = x\) and \(v = x + 1\).
- Then, \(u' = 1\) and \(v' = 1\).
- \( y' = \frac{(x+1) \cdot 1 - x \cdot 1}{(x+1)^2} = \frac{1}{(x+1)^2} \)
Function Differentiation
Understanding function differentiation is crucial in calculus. Differentiation measures how a function changes as its input changes. It’s all about finding the rate of change or the slope of the function’s graph.
For practical purposes, when you have a function like the one above \(y = \frac{x}{x+1}\), it might help to break it into parts.
Differentiation rules like the power rule, product rule, and, as mentioned previously, the quotient rule, are key to handling a wide array of functions.
For complex functions, sometimes it’s necessary to manipulate or simplify the function—for example, getting terms over a common denominator—to make differentiation easier. This preparation step allows you to apply differentiation rules more straightforwardly.
For practical purposes, when you have a function like the one above \(y = \frac{x}{x+1}\), it might help to break it into parts.
- You have a numerator \(x\) and a denominator \(x+1\).
Differentiation rules like the power rule, product rule, and, as mentioned previously, the quotient rule, are key to handling a wide array of functions.
For complex functions, sometimes it’s necessary to manipulate or simplify the function—for example, getting terms over a common denominator—to make differentiation easier. This preparation step allows you to apply differentiation rules more straightforwardly.
Simplifying Expressions
Simplifying expressions is a fundamental skill in calculus that makes differentiation and, in fact, many other operations, much easier.
When you start with a complex expression, like the original \( y = \frac{1}{1 + \frac{1}{x}} \), converting it into a simpler form can be incredibly beneficial.
For instance, by rewriting it, you find a common denominator:
Simplification often uncovers hidden relationships or patterns that are not immediately visible.It’s a step-by-step process:
When you start with a complex expression, like the original \( y = \frac{1}{1 + \frac{1}{x}} \), converting it into a simpler form can be incredibly beneficial.
For instance, by rewriting it, you find a common denominator:
- Convert \( \frac{1}{1 + \frac{1}{x}} \) to \( \frac{x}{x+1} \)
Simplification often uncovers hidden relationships or patterns that are not immediately visible.It’s a step-by-step process:
- Identify parts of the expression that can be rewritten.
- Factor, combine like terms, or find a common denominator as needed.