Textbook Question
Derivative Calculations
In Exercises 1–8, given y = f(u) and u = g(x), find dy/dx = f'(g(x)) g'(x).
y = cos u, u = −x/3
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 3.6.63
Second Derivatives
Find y'' in Exercises 59–64.
y = x(2x + 1)⁴
Verified step by step guidance1
First, identify the function y = x(2x + 1)^4. We need to find the second derivative y''.
To find the first derivative y', apply the product rule. The product rule states that if you have a function y = u(x)v(x), then y' = u'(x)v(x) + u(x)v'(x). Here, let u(x) = x and v(x) = (2x + 1)^4.
Calculate the derivatives: u'(x) = 1 and v'(x) using the chain rule. For v(x) = (2x + 1)^4, let w = 2x + 1, then v(x) = w^4. The derivative v'(x) = 4w^3 * (dw/dx) = 4(2x + 1)^3 * 2.
Substitute the derivatives back into the product rule: y' = 1 * (2x + 1)^4 + x * 8(2x + 1)^3.
Simplify y' and then find the second derivative y'' by differentiating y' again. Use the product rule and chain rule as necessary to find y''.
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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 differentiation rule used when finding the derivative of a product of two functions. If you have two functions u(x) and v(x), the derivative of their product is given by u'(x)v(x) + u(x)v'(x). This rule is essential for differentiating expressions like y = x(2x + 1)⁴, where x is multiplied by another function.
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The Product Rule
Chain Rule
The chain rule is used to differentiate composite functions, where one function is nested inside another. If you have a function y = f(g(x)), the derivative is f'(g(x)) * g'(x). In the given problem, the chain rule helps differentiate the term (2x + 1)⁴, treating it as a composite function with an inner function g(x) = 2x + 1.
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05:02Intro to the Chain Rule
Second Derivative
The second derivative, denoted as y'', is the derivative of the first derivative y'. It provides information about the curvature or concavity of the function's graph. Calculating the second derivative involves differentiating the first derivative, often requiring the application of differentiation rules like the product and chain rules multiple times.
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06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.
f(x) = { 2x − 1, x ≥ 0
x² + 2x + 7, x < 0
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = x⁻¹/² sec (2x)²
Textbook Question
Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.
Textbook Question
Find the derivatives of the functions in Exercises 17–28.
y = ((x + 1)(x + 2)) / ((x − 1)(x − 2))
Textbook Question
Derivatives
In Exercises 1–18, find dy/dx.
y = (sec x + tan x)(sec x − tan x)