Textbook Question
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
g(x) = 8 / x², (2, 2)
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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 3.6.70
Finding Derivative Values
In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.
f(u) = u + 1/cos²u, u = g(x) = πx, x = 1/4
Verified step by step guidance1
First, understand that you need to find the derivative of the composite function (f ∘ g)(x), which is f(g(x)).
Apply the chain rule for derivatives, which states that (f ∘ g)'(x) = f'(g(x)) * g'(x).
Calculate g'(x) for g(x) = πx. Since g(x) is a linear function, g'(x) = π.
Next, find f'(u) for f(u) = u + 1/cos²u. Use the derivative rules: the derivative of u is 1, and for 1/cos²u, apply the chain rule and the derivative of trigonometric functions.
Evaluate (f ∘ g)'(x) at x = 1/4 by substituting g(1/4) into f'(u) and multiplying by g'(1/4).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. If you have a function h(x) = f(g(x)), the derivative h'(x) is found by multiplying the derivative of the outer function f at g(x) by the derivative of the inner function g at x. This is essential for finding (f ∘ g)'(x).
Recommended video:
05:02
Intro to the Chain Rule
Derivative of Trigonometric Functions
Understanding how to differentiate trigonometric functions is crucial. For example, the derivative of cos(u) is -sin(u), and using the chain rule, the derivative of 1/cos²(u) involves applying the power rule and the derivative of cosine. This knowledge is necessary to differentiate f(u) = u + 1/cos²(u).
Recommended video:
6:04Introduction to Trigonometric Functions
Substitution
Substitution involves replacing variables with given values to simplify the differentiation process. In this problem, you substitute u = g(x) = πx and x = 1/4 into the functions. This step is crucial for evaluating the derivative at the specific point x = 1/4, ensuring accurate computation of (f ∘ g)'(x).
Recommended video:
04:27Substitution With an Extra Variable
Related Practice
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
f(x) = √(7 + x sec x)
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Textbook Question
Derivatives in Differential Form
In Exercises 17–28, find dy.
2y³/² + xy − x = 0
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Textbook Question
Assume that f'(3) = −1, g'(2) = 5, g(2) = 3, and y = f(g(x)). What is y' at x = 2?
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Textbook Question
In Exercises 9–18, write the function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x.
y = (4 − 3x)⁹
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Textbook Question
Computer Explorations
Use a CAS to perform the following steps in Exercises 55–62.
a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.
xy³ + tan(x + y) = 1, P(π/4, 0)
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