Textbook Question
The Reciprocal Rule
b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.57b
Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.
x ƒ(x) g(x) ƒ'(x) g'(x)
0 1 1 -3 1/2
1 3 5 1/2 -4
Find the first derivatives of the following combinations at the given value of x.
b. ƒ(x)g²(x), x = 0
Verified step by step guidance1
To find the derivative of the function ƒ(x)g²(x) at x = 0, we need to apply the product rule and the chain rule. The product rule states that the derivative of a product of two functions u(x) and v(x) is given by u'(x)v(x) + u(x)v'(x).
In this case, let u(x) = ƒ(x) and v(x) = g²(x). We need to find the derivative of v(x) using the chain rule. The chain rule states that the derivative of a composite function h(g(x)) is h'(g(x))g'(x).
Since v(x) = g²(x), we can express it as h(g(x)) where h(x) = x². The derivative h'(x) is 2x, so the derivative of v(x) = g²(x) is 2g(x)g'(x).
Now, apply the product rule: the derivative of ƒ(x)g²(x) is ƒ'(x)g²(x) + ƒ(x)(2g(x)g'(x)).
Substitute the given values at x = 0: ƒ'(0) = -3, g(0) = 1, g'(0) = 1/2, and ƒ(0) = 1. Calculate ƒ'(0)g²(0) + ƒ(0)(2g(0)g'(0)) using these values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Product Rule
The Product Rule is a fundamental principle in calculus used to differentiate products of two functions. It states that if you have two functions, ƒ(x) and g(x), the derivative of their product is given by ƒ'(x)g(x) + ƒ(x)g'(x). This rule is essential for finding the derivative of combinations of functions, such as ƒ(x)g²(x), as it allows us to apply the differentiation process systematically.
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The Product Rule
Chain Rule
The Chain Rule is another critical differentiation technique used when dealing with composite functions. It states that if a function y is composed of another function u, then the derivative of y with respect to x is the derivative of y with respect to u multiplied by the derivative of u with respect to x. In the context of g²(x), the Chain Rule helps us differentiate the square of the function g(x) effectively.
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05:02Intro to the Chain Rule
Evaluating Derivatives at Specific Points
Evaluating derivatives at specific points involves substituting the given x-value into the derivative expression to find the slope of the function at that point. In this problem, we need to compute the derivatives of ƒ(x)g²(x) at x = 0, which requires using the values of ƒ(0), g(0), ƒ'(0), and g'(0) provided in the table. This step is crucial for obtaining the final numerical result.
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04:50Critical Points
Related Practice
Textbook Question
Particle motion At time t ≥ 0, the velocity of a body moving along the horizontal s-axis is v = t² − 4t + 3.
b. When is the body moving forward? Backward?
Textbook Question
Theory and Examples
In Exercises 51–54,
b. Graph y = f(x) and y = f'(x) side by side using separate sets of coordinate axes, and answer the following questions.
y = x⁴/4
Textbook Question
Average single-family home prices P (in thousands of dollars) in Sacramento, California, are shown in the accompanying figure from the beginning of 2006 through the end of 2015.
b. Estimate home prices at the end of
i) 2007 ii) 2012 iii) 2015
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Textbook Question
Common linear approximations at x = 0 Find the linearizations of the following functions at x = 0.
b. cos x
Textbook Question
Recovering a function from its derivative
b. Repeat part (a), assuming that the graph starts at (−2, 0) instead of (−2, 3).