Textbook Question
Derivatives
In Exercises 27–32, find dp/dq.
p = (sin q + cos q) / cos q
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 3.6.69
Finding Derivative Values
In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.
f(u) = cot(πu/10), u = g(x) = 5√x, x = 1
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)). This involves using the chain rule for differentiation.
Identify the functions involved: f(u) = cot(πu/10) and u = g(x) = 5√x. You need to differentiate f with respect to u and g with respect to x.
Apply the chain rule: (f ∘ g)'(x) = f'(g(x)) * g'(x). This means you need to find f'(u) and g'(x) and then multiply them.
Differentiate f(u) = cot(πu/10) with respect to u. Use the derivative of cotangent, which is -csc²(u), and apply the chain rule to account for the π/10 factor.
Differentiate g(x) = 5√x with respect to x. Recall that √x can be expressed as x^(1/2), and use the power rule to find g'(x).
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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 with respect to its inner function g, by the derivative of the inner function g with respect to x. This is expressed as h'(x) = f'(g(x)) * g'(x).
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05:02
Intro to the Chain Rule
Derivative of Cotangent Function
The derivative of the cotangent function, cot(u), with respect to u is -csc^2(u). This derivative is crucial when differentiating functions involving cotangent, as it allows us to apply the chain rule effectively. In the context of the given problem, understanding this derivative helps in finding the derivative of f(u) = cot(πu/10).
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5:37Introduction to Cotangent Graph
Square Root Function Derivative
The derivative of the square root function, √x, is (1/2)x^(-1/2). This derivative is essential when dealing with functions that involve square roots, such as g(x) = 5√x. Knowing this allows us to compute g'(x), which is necessary for applying the chain rule to find the derivative of the composite function (f ∘ g)(x).
Recommended video:
01:32Derivatives of Other Trig Functions Example 1
Related Practice
Textbook Question
Differential Estimates of Change
In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.
The change in the volume V = (4/3)πr³ of a sphere when the radius changes from r₀ to r₀ + dr
Textbook Question
Derivatives
In Exercises 23–26, find dr/dθ.
r = θ sin θ + cos θ
Textbook Question
If r = sin(f(t)), f(0) = π/3, and f'(0) = 4, then what is dr/dt at t = 0?
Textbook Question
Moving along a parabola A particle moves along the parabola y = x² in the first quadrant in such a way that its x-coordinate (measured in meters) increases at a steady 10 m/sec. How fast is the angle of inclination θ of the line joining the particle to the origin changing when x = 3 m?
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = (θ² + sec θ + 1)³