Find the second derivative of the function f(t)= t³sect. Of"(t)= t³sect(20+tsect+10ttant+ftan²t) Of"(t) = f'sect(20+fsect+10ttant+ftan²t) O f"(t) = f'sect(20+sect+ 5ttant+ttan²t) ○ _fƒ"(t)=1²³sect(20+²sec²t+10ttant+ftant) O f"(t)= t³sect(20+²sec²t+10ttant + ttan²t)

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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**Problem:**
Find the second derivative of the function \( f(t) = t^5 \sec t \).

**Options:**
1. \( f''(t) = t^3 \sec \left( 20 + t \sec^2 t + 10 \tan t + t^2 \tan^2 t \right) \)
2. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 10 \tan t + t^2 \tan^2 t \right) \)
3. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 5 \tan t + t^2 \tan^2 t \right) \)
4. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 10 \tan t + t^2 \tan \right) \)
5. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 10 \tan t + t \tan^2 t \right) \)

Select the correct option for the second derivative of the given function.
Transcribed Image Text:**Problem:** Find the second derivative of the function \( f(t) = t^5 \sec t \). **Options:** 1. \( f''(t) = t^3 \sec \left( 20 + t \sec^2 t + 10 \tan t + t^2 \tan^2 t \right) \) 2. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 10 \tan t + t^2 \tan^2 t \right) \) 3. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 5 \tan t + t^2 \tan^2 t \right) \) 4. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 10 \tan t + t^2 \tan \right) \) 5. \( f''(t) = t^3 \sec \left( 20 + t^2 \sec^2 t + 10 \tan t + t \tan^2 t \right) \) Select the correct option for the second derivative of the given function.
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