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...
Related questions
Question
![**Problem 30:**
Given the function \( f(t) = \sec t \), find the second derivative \( f''(\pi/4) \).
**Solution Approach:**
1. **Find the First Derivative**:
The derivative of \( \sec t \) with respect to \( t \) is \( f'(t) = \sec t \tan t \).
2. **Find the Second Derivative**:
Differentiate \( f'(t) = \sec t \tan t \) with respect to \( t \) to obtain \( f''(t) \).
3. **Evaluate at \( t = \pi/4 \)**:
Substitute \( t = \pi/4 \) into \( f''(t) \) to find \( f''(\pi/4) \).
**Calculation Steps**:
- First, find the derivative of \( \sec t \):
\[
f'(t) = \sec t \tan t
\]
- Then, differentiate \( f'(t) \) to find \( f''(t) \):
\[
f''(t) = \frac{d}{dt}(\sec t \tan t)
\]
- Apply the product rule: if \( u = \sec t \) and \( v = \tan t \), then:
\[
f''(t) = u'v + uv' = (\sec t \tan t)\tan t + (\sec t)(\sec^2 t)
\]
Simplifying further:
\[
f''(t) = \sec t \tan^2 t + \sec^3 t
\]
- Evaluate at \( t = \pi/4 \):
- \(\sec(\pi/4) = \sqrt{2}\)
- \(\tan(\pi/4) = 1\)
Substitute into \( f''(t) \):
\[
f''(\pi/4) = \sqrt{2} \cdot 1^2 + (\sqrt{2})^3
\]
\[
f''(\pi/4) = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}
\]
Thus, the second derivative of the function at \( t = \pi/](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90d2137b-83f2-4c15-8300-1a033b76054d%2F45d2c1a0-544a-4b85-a53b-8da78a1b0bfe%2Feotycx_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 30:**
Given the function \( f(t) = \sec t \), find the second derivative \( f''(\pi/4) \).
**Solution Approach:**
1. **Find the First Derivative**:
The derivative of \( \sec t \) with respect to \( t \) is \( f'(t) = \sec t \tan t \).
2. **Find the Second Derivative**:
Differentiate \( f'(t) = \sec t \tan t \) with respect to \( t \) to obtain \( f''(t) \).
3. **Evaluate at \( t = \pi/4 \)**:
Substitute \( t = \pi/4 \) into \( f''(t) \) to find \( f''(\pi/4) \).
**Calculation Steps**:
- First, find the derivative of \( \sec t \):
\[
f'(t) = \sec t \tan t
\]
- Then, differentiate \( f'(t) \) to find \( f''(t) \):
\[
f''(t) = \frac{d}{dt}(\sec t \tan t)
\]
- Apply the product rule: if \( u = \sec t \) and \( v = \tan t \), then:
\[
f''(t) = u'v + uv' = (\sec t \tan t)\tan t + (\sec t)(\sec^2 t)
\]
Simplifying further:
\[
f''(t) = \sec t \tan^2 t + \sec^3 t
\]
- Evaluate at \( t = \pi/4 \):
- \(\sec(\pi/4) = \sqrt{2}\)
- \(\tan(\pi/4) = 1\)
Substitute into \( f''(t) \):
\[
f''(\pi/4) = \sqrt{2} \cdot 1^2 + (\sqrt{2})^3
\]
\[
f''(\pi/4) = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}
\]
Thus, the second derivative of the function at \( t = \pi/
Expert Solution
Step 1
Topic- Double derivative of a function.
we are given
f(t)=sec t
and we know that
Step by step
Solved in 2 steps
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