Textbook Question
In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.
g(x) = { x²/³, x ≥ 0
x¹/³, x < 0
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 3.21
Find the derivatives of the functions in Exercises 1–42.
𝔂 = 1 x² csc 2
2 x
Verified step by step guidance1
Step 1: Identify the function components. The given function is \( y = \frac{1}{x^2} \csc(2x) \). This is a product of two functions: \( u(x) = \frac{1}{x^2} \) and \( v(x) = \csc(2x) \).
Step 2: Apply the product rule for differentiation. The product rule states that if \( y = u(x) \cdot v(x) \), then \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).
Step 3: Differentiate \( u(x) = \frac{1}{x^2} \). Use the power rule: \( u'(x) = -2x^{-3} \).
Step 4: Differentiate \( v(x) = \csc(2x) \). Use the chain rule: \( v'(x) = -2 \csc(2x) \cot(2x) \).
Step 5: Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the product rule formula to find \( y' \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivatives
The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the curve of the function at any given point. Derivatives can be computed using various rules, such as the power rule, product rule, and quotient rule, depending on the form of the function.
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Derivatives
Cosecant Function (csc)
The cosecant function, denoted as csc(x), is the reciprocal of the sine function, defined as csc(x) = 1/sin(x). It is important in calculus when dealing with trigonometric functions, especially when finding derivatives. Understanding how to differentiate csc(x) and its properties is essential for solving problems involving trigonometric derivatives.
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Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the ratio of two other functions. If you have a function defined as f(x) = g(x)/h(x), the derivative is given by f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))². This rule is crucial when differentiating functions that involve division, such as the one presented in the question.
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Related Practice
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = x³ - 3 (x² + π²)
1
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Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x cos(2x + 3y) = y sin x
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 lateral surface area S = πr√(r² + h²) of a right circular cone when the radius changes from r₀ to r₀ + dr and the height does not change
Textbook Question
Slopes, Tangent Lines, and Normal Lines
In Exercises 31–40, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
x²y² = 9, (–1,3)
Textbook Question
Is there a value of b that will make
g(x) = { x + b, x < 0
cos x, x ≥ 0
continuous at x = 0? Differentiable at x = 0? Give reasons for your answers.