Problem 22
Question
Match the function with the rule that you would use to find the derivative most efficiently. \(\begin{array}{ll}{\text { (a) Simple Power Rule }} & {\text { (b) Constant Rule }} \\ {\text { (c) General Power Rule }} & {\text { (d) Quotient Rule }}\end{array}\) $$ f(x)=\frac{5}{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
(d) Quotient Rule
1Step 1: Analyze the Function Form
First, take a close look at the function \(f(x) = \frac{5}{x^{2}+1}\). This is a function that is a ratio of a constant (the numerator is 5, a constant) and a polynomial (the denominator is \(x^{2}+1\), a polynomial).
2Step 2: Analyze Given Rules
Now take a look at the rules. The Simple Power Rule is used when we are dealing with functions of the form \(cx^n\), where c is a constant and n is a real number. The Constant Rule is used when our function is a constant c. The General Power Rule is used when our function is a composition of two functions of the form \(f(g(x))\), where f and g are both power functions, i.e., they can be written in the form \(cx^n\). Lastly, the Quotient Rule is used when our function is the ratio of two other functions.
3Step 3: Identify the Rule
From the analysis of the function and the rules, it is clear that the function fits the conditions for the Quotient Rule. The Quotient Rule is applied when we have a function that is the division of two other functions (in this case, the constant 5 and the polynomial \(x^{2}+1\)). Hence, the Quotient Rule is the most efficient rule for this problem.
Key Concepts
DerivativeQuotient RuleGeneral Power Rule
Derivative
In calculus, the derivative represents a fundamental concept that describes the rate at which a function is changing at any given point. It is essentially a tool used to find the slope of a function's curve at a specific point. More formally, the derivative of a function at a given point is defined as the limit of the average rate of change of the function as the change in input approaches zero. This gives us an instantaneous rate of change.
To differentiate a function effectively, different rules are applied based on the function's structure. These rules include the power rule, the constant rule, the quotient rule, among others. Each rule simplifies the process of finding the derivative, allowing us to handle different types of functions with ease.
To differentiate a function effectively, different rules are applied based on the function's structure. These rules include the power rule, the constant rule, the quotient rule, among others. Each rule simplifies the process of finding the derivative, allowing us to handle different types of functions with ease.
- The power rule applies to functions of the form \( cx^n \).
- The constant rule is used for constant functions.
- The quotient rule tackles functions presented as a division between two other functions.
Quotient Rule
The quotient rule is specifically useful when dealing with functions that are expressed as one function divided by another. This rule makes handling derivatives of fractions straightforward, especially when you have functions that are not easily simplified.
Mathematically, the quotient rule is stated as follows: If you have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable functions, the derivative \( f'(x) \) can be calculated using:\[ f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}\]This formula comes from the product and chain rule applications and requires both the numerator and denominator to be differentiable. To successfully use the quotient rule, ensure that the denominator \( v(x) \) is never zero to avoid undefined behavior.
Mathematically, the quotient rule is stated as follows: If you have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable functions, the derivative \( f'(x) \) can be calculated using:\[ f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}\]This formula comes from the product and chain rule applications and requires both the numerator and denominator to be differentiable. To successfully use the quotient rule, ensure that the denominator \( v(x) \) is never zero to avoid undefined behavior.
- "\( u(x) \)" refers to the numerator.
- "\( v(x) \)" refers to the denominator.
- The subtraction order in the numerator \( v(x) \cdot u'(x) - u(x) \cdot v'(x) \) is crucial, ensuring accuracy in calculation.
General Power Rule
The general power rule is an extension of the simple power rule. It is particularly valuable when differentiating composite functions, which involve compositions like \( f(g(x)) \). Here, both \( f \) and \( g \) can be power functions or have their structures.
The general power rule can be applied as follows: If you have a function \( y = [g(x)]^n \), the derivative \( y' \) is determined by:\[ y'(x) = n[g(x)]^{n-1} \, g'(x)\]This rule underscores the importance of applying both the power and chain rules, making it well-suited to handle more complex functions. When implementing the general power rule, pay attention to the internal function \( g(x) \) and its derivative \( g'(x) \), which play crucial roles.
The general power rule can be applied as follows: If you have a function \( y = [g(x)]^n \), the derivative \( y' \) is determined by:\[ y'(x) = n[g(x)]^{n-1} \, g'(x)\]This rule underscores the importance of applying both the power and chain rules, making it well-suited to handle more complex functions. When implementing the general power rule, pay attention to the internal function \( g(x) \) and its derivative \( g'(x) \), which play crucial roles.
- \( n \) defines the power to which the entire function is raised.
- The derivative \( g'(x) \) is essential, capturing how the internal function changes.
- This technique streamlines what would otherwise require multiple applications of simpler rules.
Other exercises in this chapter
Problem 22
Advertising costs A retail sporting goods store estimates that weekly sales \(S\) and weekly advertising costs \(x\) are related by the equation \(S=2250+50 x+0
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Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the given point. Equation \(\quad\) Point \(\sqrt{x y}=x-2 y\) \(\quad\) \((4,1)\)
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find the third derivative of the function. $$ f(x)=\frac{1}{x} $$
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Use the limit definition to find the slope of the tangent line to the graph of \(f\) at the given point. $$ f(x)=x^{3}+2 x ;(1,3) $$
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