d. (e sec z) = ? da O e sec z (1 – tan z) O e sec z (1 + tan 2) O e (sec a + tan a) O e (sec z tan z). Question 5 Suppose f (27/3) = -6 and f' (2n/3) = 4, and let g(z) = f(x) cosT. Find g'(27/3).
d. (e sec z) = ? da O e sec z (1 – tan z) O e sec z (1 + tan 2) O e (sec a + tan a) O e (sec z tan z). Question 5 Suppose f (27/3) = -6 and f' (2n/3) = 4, and let g(z) = f(x) cosT. Find g'(27/3).
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 4
Find the derivative with respect to \(x\):
\[
\frac{d}{dx} \left( e^x \sec x \right) = ?
\]
Options:
- \( e^x \sec x (1 - \tan x) \)
- \( e^x \sec x (1 + \tan x) \)
- \( e^x (\sec x + \tan^2 x) \)
- \( e^x (\sec x - \tan^2 x) \)
### Question 5
Suppose \( f(2\pi/3) = -6 \) and \( f'(2\pi/3) = 4 \), and let \( g(x) = f(x) \cos x \). Find \( g'(2\pi/3) \).
Options:
- \( 2 - 3\sqrt{3} \)
- \( 2 + 3\sqrt{3} \)
- \( 2 - 3\sqrt{3} \)
- \( 2 + 3\sqrt{3} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08f6f3ae-62d0-4ecf-a7d1-a70a7054db3c%2F2a379d66-5efd-46c5-8fa0-7cbb72009302%2Ff7vc68v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 4
Find the derivative with respect to \(x\):
\[
\frac{d}{dx} \left( e^x \sec x \right) = ?
\]
Options:
- \( e^x \sec x (1 - \tan x) \)
- \( e^x \sec x (1 + \tan x) \)
- \( e^x (\sec x + \tan^2 x) \)
- \( e^x (\sec x - \tan^2 x) \)
### Question 5
Suppose \( f(2\pi/3) = -6 \) and \( f'(2\pi/3) = 4 \), and let \( g(x) = f(x) \cos x \). Find \( g'(2\pi/3) \).
Options:
- \( 2 - 3\sqrt{3} \)
- \( 2 + 3\sqrt{3} \)
- \( 2 - 3\sqrt{3} \)
- \( 2 + 3\sqrt{3} \)
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