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
![### Calculus Problem
**Objective:**
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
**Problem Statement:**
\[
y = \int_{1}^{\cos(x)} (3 + v^6)^5 \, dv
\]
**Task:**
Calculate the derivative \( y' \).
**Solution:**
To find the derivative \( y' \) using Part 1 of the Fundamental Theorem of Calculus, you should remember that if:
\[
y = \int_{a}^{g(x)} f(v) \, dv
\]
then the derivative is given by:
\[
y' = f(g(x)) \cdot g'(x)
\]
Applying this logic to the given integral:
1. **Function Inside the Integral:** \( f(v) = (3 + v^6)^5 \)
2. **Upper Limit of Integration:** \( g(x) = \cos(x) \)
3. **Derivative of the Upper Limit:** \( g'(x) = -\sin(x) \)
Thus, the derivative \( y' \) is:
\[
y' = (3 + (\cos(x))^6)^5 \cdot (-\sin(x))
\]
Remember to multiply the function evaluated at \( \cos(x) \) by the derivative of \( \cos(x) \), which is \( -\sin(x) \).
**Final Answer:**
\[
y' = -(3 + (\cos(x))^6)^5 \sin(x)
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66f2d714-57ab-40fc-941f-d13524a78f49%2F8137c1ce-f425-4e31-8ac3-e38f8505ddde%2F2tqjf3_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem
**Objective:**
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
**Problem Statement:**
\[
y = \int_{1}^{\cos(x)} (3 + v^6)^5 \, dv
\]
**Task:**
Calculate the derivative \( y' \).
**Solution:**
To find the derivative \( y' \) using Part 1 of the Fundamental Theorem of Calculus, you should remember that if:
\[
y = \int_{a}^{g(x)} f(v) \, dv
\]
then the derivative is given by:
\[
y' = f(g(x)) \cdot g'(x)
\]
Applying this logic to the given integral:
1. **Function Inside the Integral:** \( f(v) = (3 + v^6)^5 \)
2. **Upper Limit of Integration:** \( g(x) = \cos(x) \)
3. **Derivative of the Upper Limit:** \( g'(x) = -\sin(x) \)
Thus, the derivative \( y' \) is:
\[
y' = (3 + (\cos(x))^6)^5 \cdot (-\sin(x))
\]
Remember to multiply the function evaluated at \( \cos(x) \) by the derivative of \( \cos(x) \), which is \( -\sin(x) \).
**Final Answer:**
\[
y' = -(3 + (\cos(x))^6)^5 \sin(x)
\]
Expert Solution
Step 1
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning