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dt dp dt о dp dt о dp dt S 3t tan 3t = 3sec t tan 3t 3sec3t tan 3t = 3 sec 3t

dt dp dt о dp dt о dp dt S 3t tan 3t = 3sec t tan 3t 3sec3t tan 3t = 3 sec 3t

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
**Find the Derivative: \( p = \sec 3t \)** Select the correct derivative of the function \( p = \sec 3t \) with respect to \( t \). Options: - \( \frac{dp}{dt} = 3t \tan 3t \) - \( \frac{dp}{dt} = 3 \sec t \tan 3t \) (selected) - \( \frac{dp}{dt} = 3 \sec 3t \tan 3t \) - \( \frac{dp}{dt} = 3 \sec 3t \) **Explanation:** This question presents four options for the derivative of the function \( p = \sec 3t \). The selected option is the second one, \( \frac{dp}{dt} = 3 \sec t \tan 3t \). However, note that typically the derivative of \( \sec 3t \) is \( 3 \sec 3t \tan 3t \) due to the chain rule.
Transcribed Image Text:**Find the Derivative: \( p = \sec 3t \)** Select the correct derivative of the function \( p = \sec 3t \) with respect to \( t \). Options: - \( \frac{dp}{dt} = 3t \tan 3t \) - \( \frac{dp}{dt} = 3 \sec t \tan 3t \) (selected) - \( \frac{dp}{dt} = 3 \sec 3t \tan 3t \) - \( \frac{dp}{dt} = 3 \sec 3t \) **Explanation:** This question presents four options for the derivative of the function \( p = \sec 3t \). The selected option is the second one, \( \frac{dp}{dt} = 3 \sec t \tan 3t \). However, note that typically the derivative of \( \sec 3t \) is \( 3 \sec 3t \tan 3t \) due to the chain rule.
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