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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.35
Chapter 3, Problem 3.35

Find the derivatives of the functions in Exercises 1–42.


𝓻 = ( sin θ )²
( cos θ - 1 )

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1
Identify the function for which you need to find the derivative. Here, the function is \( r(\theta) = (\sin \theta)^2 (\cos \theta - 1) \).
Apply the product rule for differentiation, which states that if you have a function \( u(\theta) \cdot v(\theta) \), its derivative is \( u'(\theta) \cdot v(\theta) + u(\theta) \cdot v'(\theta) \). In this case, let \( u(\theta) = (\sin \theta)^2 \) and \( v(\theta) = (\cos \theta - 1) \).
Differentiate \( u(\theta) = (\sin \theta)^2 \) using the chain rule. The chain rule states that the derivative of \( (f(g(\theta))) \) is \( f'(g(\theta)) \cdot g'(\theta) \). Here, \( f(x) = x^2 \) and \( g(\theta) = \sin \theta \). So, \( u'(\theta) = 2 \sin \theta \cdot \cos \theta \).
Differentiate \( v(\theta) = (\cos \theta - 1) \). The derivative of \( \cos \theta \) is \( -\sin \theta \), and the derivative of a constant is 0. Therefore, \( v'(\theta) = -\sin \theta \).
Substitute \( u'(\theta) \), \( u(\theta) \), \( v'(\theta) \), and \( v(\theta) \) into the product rule formula: \( r'(\theta) = [2 \sin \theta \cdot \cos \theta] \cdot (\cos \theta - 1) + (\sin \theta)^2 \cdot [-\sin \theta] \). Simplify the expression to find the derivative.

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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 at which a function changes at any given point. It is a fundamental concept in calculus that provides information about the slope of the tangent line to the curve of the function. Derivatives can be computed using various rules, such as the power rule, product rule, and chain rule, depending on the form of the function.
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Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic functions that relate angles to ratios of sides in right triangles. In calculus, these functions are essential for modeling oscillatory behavior and are frequently encountered in derivatives. Understanding their properties, such as their derivatives (e.g., the derivative of sin θ is cos θ), is crucial for solving problems involving these functions.
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Polar Coordinates

Polar coordinates provide a way to represent points in a plane using a distance from a reference point and an angle from a reference direction. In the context of the given function, r = (sin θ)² / (cos θ - 1), understanding how to differentiate functions expressed in polar coordinates is important. The conversion between polar and Cartesian coordinates may also be necessary for certain derivative calculations.
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Related Practice
Textbook Question

Assume that y = 5x and dx/dt = 2. Find dy/dt

Textbook Question

Derivatives


In Exercises 1–18, find dy/dx.


f(x) = x³ sin x cos x

Textbook Question

In Exercises 41–58, find dy/dt.


y = 4 sin(√(1 + √t))

Textbook Question

Derivatives


In Exercises 27–32, find dp/dq.


p = (q sin q) / (q² − 1)

Textbook Question

Finding Derivative Values


In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.


f(u) = 1 − (1/u), u = g(x) = (1 / (1 − x)), x = −1

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Textbook Question

Find the first and second derivatives of the functions in Exercises 33–38.


s = (t² + 5t − 1) / t²