Chapter 23: Problem 37
Find the derivative of \(y=1 / x^{3}\) as (a) a quotient and (b) a negative power of \(x\) and show that the results are the same.
Short Answer
Expert verified
The derivative is \(-\frac{3}{x^4}\) or \(-3x^{-4}\), both methods give the same result.
Step by step solution
01
Identify the Function as a Quotient
The given function is \(y = \frac{1}{x^3}\). This is a quotient where the numerator is 1 and the denominator is \(x^3\).
02
Differentiate Using the Quotient Rule
The quotient rule states that if \( y = \frac{u}{v} \), then the derivative \( y' \) is given by \( \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u = 1 \) and \( v = x^3 \). \[ u' = 0, \quad v' = 3x^2 \] Using the quotient rule: \[ y' = \frac{x^3 \cdot 0 - 1 \cdot 3x^2}{(x^3)^2} = \frac{-3x^2}{x^6} = -\frac{3}{x^4} \]
03
Rewrite as a Power of x with a Negative Exponent
Rewrite the function as a power of \( x \): \( y = x^{-3} \). This form uses a negative exponent to allow easier differentiation using power rules.
04
Differentiate Using the Power Rule
The power rule states that the derivative \( y' = nx^{n-1} \). For \( y = x^{-3} \), \[ y' = -3x^{-3-1} = -3x^{-4} \]
05
Verify Same Result with Different Approaches
Both derivatives from the quotient rule \( -\frac{3}{x^4} \) and the power rule \( -3x^{-4} \) are equivalent, showing consistency in the methods.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
When you have a function that is expressed as a division of two functions, like \( y = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \), you can use the quotient rule to find its derivative. The quotient rule helps us find the rate of change of a quotient of two functions. This is especially useful when both the numerator and denominator are variable-dependent expressions.
The formula for the quotient rule is:
For the function \( y = \frac{1}{x^3} \), we can identify:
The formula for the quotient rule is:
- \( y' = \frac{v \cdot u' - u \cdot v'}{v^2} \)
For the function \( y = \frac{1}{x^3} \), we can identify:
- \( u = 1 \); hence, \( u' = 0 \)
- \( v = x^3 \); therefore, \( v' = 3x^2 \)
- \( y' = \frac{x^3 \cdot 0 - 1 \cdot 3x^2}{(x^3)^2} = \frac{-3x^2}{x^6} = -\frac{3}{x^4} \)
Power Rule
The power rule is perhaps one of the most straightforward rules in calculus for finding derivatives. It applies to functions in the form of \( x^n \), where \( n \) is any real number. This rule is incredibly useful because it simplifies the differentiation process, allowing you to handle exponents easily without any need of a second equation or function.
The power rule states:
The power rule states:
- If \( y = x^n \) then the derivative \( y' = nx^{n-1} \)
- \( y' = -3x^{-3-1} = -3x^{-4} \)
Negative Exponents
A negative exponent represents the reciprocal of the base raised to the opposite positive power. This concept is pivotal when working with functions like \( y = \frac{1}{x^3} \), which can be rewritten as \( y = x^{-3} \). Understanding negative exponents allows for easy manipulation and differentiation of complex functions.
Consider the expression \( x^{-n} \):
In our exercise:
Consider the expression \( x^{-n} \):
- \( x^{-n} = \frac{1}{x^n} \)
In our exercise:
- The function \( y = x^{-3} \) can be differentiated using the power rule to yield \( y' = -3x^{-4} \)