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
![**Problem Statement:**
Find the rate of change of \( y \) with respect to \( x \) at the indicated value of \( x \).
\[
y = \frac{\sin(x)}{5 - \cos(x)}; \quad x = \frac{\pi}{2}
\]
\( y' = \) [blank for the answer]
**Explanation:**
We are given the function \( y = \frac{\sin(x)}{5 - \cos(x)} \) and need to find its derivative with respect to \( x \) at \( x = \frac{\pi}{2} \).
**Solution Steps:**
1. **Differentiate the function \( y \) using the quotient rule**, which states that if \( y = \frac{u(x)}{v(x)} \), then:
\[
y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
2. In our function, set \( u(x) = \sin(x) \) and \( v(x) = 5 - \cos(x) \).
3. **Calculate the derivatives**:
\[
u'(x) = \cos(x)
\]
\[
v'(x) = \sin(x)
\]
4. **Substitute into the quotient rule**:
\[
y' = \frac{\cos(x)(5 - \cos(x)) - \sin(x)(-\sin(x))}{(5 - \cos(x))^2}
\]
5. **Simplify** the expression:
\[
y' = \frac{5\cos(x) - \cos^2(x) + \sin^2(x)}{(5 - \cos(x))^2}
\]
6. **Evaluate** the derivative at \( x = \frac{\pi}{2} \):
- At \( x = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \).
\[
y' = \frac{5(0) - 0^2 + 1^2}{(5 - 0)^2} = \frac{1}{25}
\]
7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c11cb4e-496c-49a9-b15b-3a14af31a568%2F9804f0b5-ea06-4360-b2ce-9d7e4a6facf2%2Fmg4pfel_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the rate of change of \( y \) with respect to \( x \) at the indicated value of \( x \).
\[
y = \frac{\sin(x)}{5 - \cos(x)}; \quad x = \frac{\pi}{2}
\]
\( y' = \) [blank for the answer]
**Explanation:**
We are given the function \( y = \frac{\sin(x)}{5 - \cos(x)} \) and need to find its derivative with respect to \( x \) at \( x = \frac{\pi}{2} \).
**Solution Steps:**
1. **Differentiate the function \( y \) using the quotient rule**, which states that if \( y = \frac{u(x)}{v(x)} \), then:
\[
y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
2. In our function, set \( u(x) = \sin(x) \) and \( v(x) = 5 - \cos(x) \).
3. **Calculate the derivatives**:
\[
u'(x) = \cos(x)
\]
\[
v'(x) = \sin(x)
\]
4. **Substitute into the quotient rule**:
\[
y' = \frac{\cos(x)(5 - \cos(x)) - \sin(x)(-\sin(x))}{(5 - \cos(x))^2}
\]
5. **Simplify** the expression:
\[
y' = \frac{5\cos(x) - \cos^2(x) + \sin^2(x)}{(5 - \cos(x))^2}
\]
6. **Evaluate** the derivative at \( x = \frac{\pi}{2} \):
- At \( x = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \).
\[
y' = \frac{5(0) - 0^2 + 1^2}{(5 - 0)^2} = \frac{1}{25}
\]
7
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