Textbook Question
In Exercises 41–58, find dy/dt.
y = (t tan(t))¹⁰
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 3.5.26
Derivatives
In Exercises 23–26, find dr/dθ.
r = (1 + sec θ) sin θ
Verified step by step guidance1
First, identify the function r in terms of θ: r(θ) = (1 + sec(θ)) * sin(θ).
To find the derivative dr/dθ, apply the product rule, which states that if you have a function u(θ) * v(θ), then the derivative is u'(θ) * v(θ) + u(θ) * v'(θ).
Let u(θ) = 1 + sec(θ) and v(θ) = sin(θ). First, find the derivative of u(θ): u'(θ) = d/dθ [1 + sec(θ)]. The derivative of sec(θ) is sec(θ)tan(θ), so u'(θ) = sec(θ)tan(θ).
Next, find the derivative of v(θ): v'(θ) = d/dθ [sin(θ)]. The derivative of sin(θ) is cos(θ), so v'(θ) = cos(θ).
Now, apply the product rule: dr/dθ = u'(θ) * v(θ) + u(θ) * v'(θ) = [sec(θ)tan(θ) * sin(θ)] + [(1 + sec(θ)) * cos(θ)].
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivatives
A derivative represents the rate of change of a function with respect to a variable. In calculus, it is a fundamental concept that allows us to determine how a function behaves as its input changes. The derivative can be interpreted as the slope of the tangent line to the curve of the function at a given point.
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Derivatives
Parametric Equations
In this context, the equation r = (1 + sec θ) sin θ is a parametric equation where r is expressed in terms of the parameter θ. Understanding parametric equations is crucial for finding derivatives with respect to a parameter, as it involves differentiating with respect to that parameter rather than the independent variable directly.
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Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. When finding dr/dθ, the chain rule allows us to differentiate r as a function of θ, especially when r is defined in terms of other trigonometric functions. This rule is essential for correctly applying derivatives in parametric contexts.
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05:02Intro to the Chain Rule
Related Practice
Textbook Question
For what value or values of the constant m, if any, is
ƒ(x) = { sin 2x, x ≤ 0
{ mx, x > 0
a. continuous at x = 0?
b. differentiable at x = 0?
Give reasons for your answers.
Textbook Question
Derivatives in Differential Form
In Exercises 17–28, find dy.
y = x√(1 − x²)
Textbook Question
The eight curve Find the slopes of the curve y⁴ = y² – x² at the two points shown here.
Textbook Question
Are there any points on the curve y = x - 1/(2x) where the slope is 2? If so, find them.
Textbook Question
Assume that functions f and g are differentiable with f(2) = 3, f'(2) = −1, g(2) = −4, and g'(2) = 1. Find an equation of the line perpendicular to the line tangent to the graph of F(x) = (f(x) + 3) / (x − g(x)) at x = 2.