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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![Here is the transcription of the image presented in a format suitable for an educational website:
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**Problem Statement:**
**Find the derivative of the function \( g(x) = \int_{x}^{3} \cos^{2} t \, dt \).**
**Explanation:**
The function \( g(x) \) is defined as an integral with a variable lower limit \( x \) and a constant upper limit \( 3 \). To find the derivative of this function with respect to \( x \), we can apply the Fundamental Theorem of Calculus, Part 1, which states:
If \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( \frac{dF}{dx} = f(x) \).
However, since the limits are reversed in our function (from \( x \) to \( 3 \) instead of from \( 3 \) to \( x \)), we need to consider the negative sign:
\[ \frac{d}{dx} \left( \int_{x}^{3} f(t) \, dt \right) = -f(x) \]
Applying this to our function \( g(x) \):
\[ g(x) = \int_{x}^{3} \cos^{2} t \, dt \]
The integrand, \( f(t) \), in this case, is \( \cos^{2} t \). Therefore,
\[ g'(x) = - \cos^{2} x \]
Thus, the derivative of the given function is:
\[ g'(x) = - \cos^2 x \]
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This transcription is aimed at helping students understand the process of finding the derivative of a function defined by an integral with a variable limit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd65cec20-e50a-495e-bf9b-5a17b6c06d4e%2F87cf95ca-450e-4313-8fee-f1c6116486e3%2Fyxp4tnr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Here is the transcription of the image presented in a format suitable for an educational website:
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**Problem Statement:**
**Find the derivative of the function \( g(x) = \int_{x}^{3} \cos^{2} t \, dt \).**
**Explanation:**
The function \( g(x) \) is defined as an integral with a variable lower limit \( x \) and a constant upper limit \( 3 \). To find the derivative of this function with respect to \( x \), we can apply the Fundamental Theorem of Calculus, Part 1, which states:
If \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( \frac{dF}{dx} = f(x) \).
However, since the limits are reversed in our function (from \( x \) to \( 3 \) instead of from \( 3 \) to \( x \)), we need to consider the negative sign:
\[ \frac{d}{dx} \left( \int_{x}^{3} f(t) \, dt \right) = -f(x) \]
Applying this to our function \( g(x) \):
\[ g(x) = \int_{x}^{3} \cos^{2} t \, dt \]
The integrand, \( f(t) \), in this case, is \( \cos^{2} t \). Therefore,
\[ g'(x) = - \cos^{2} x \]
Thus, the derivative of the given function is:
\[ g'(x) = - \cos^2 x \]
---
This transcription is aimed at helping students understand the process of finding the derivative of a function defined by an integral with a variable limit.
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