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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![**Title: Implicit Differentiation to Find the Second Derivative**
**Problem:**
Find \( y'' \) using implicit differentiation for the equation:
\[
\cos y + \sin x = 1
\]
**Solution Steps:**
1. **Differentiate the Equation:**
Start by applying implicit differentiation to both sides of the equation with respect to \( x \):
\[
\frac{d}{dx}(\cos y) + \frac{d}{dx}(\sin x) = \frac{d}{dx}(1)
\]
2. **Applying Derivatives:**
\[
-\sin y \cdot \frac{dy}{dx} + \cos x = 0
\]
3. **Solve for the First Derivative \( y' \):**
Re-arrange to solve for \( \frac{dy}{dx} \) (denote \( y' \) as \( \frac{dy}{dx} \)):
\[
y' = \frac{\cos x}{\sin y}
\]
4. **Find the Second Derivative \( y'' \):**
Differentiate \( y' = \frac{\cos x}{\sin y} \) with respect to \( x \):
\[
y'' = \frac{d}{dx} \left( \frac{\cos x}{\sin y} \right)
\]
5. **Use the Quotient Rule:**
The quotient rule is:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]
Here, \( u = \cos x \) and \( v = \sin y \).
\[
\frac{du}{dx} = -\sin x, \quad \frac{dv}{dx} = \cos y \cdot y'
\]
Substituting into the quotient rule:
\[
y'' = \frac{\sin y (-\sin x) - \cos x (\cos y \cdot y')}{(\sin y)^2}
\]
6. **Substitute \( y' = \frac{\cos x}{\sin y} \):**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a10fbfc-d7d7-4914-b7d7-e6510260e301%2F2d198f7b-23cd-4714-b394-c288821ae777%2Fvzbgrm9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Implicit Differentiation to Find the Second Derivative**
**Problem:**
Find \( y'' \) using implicit differentiation for the equation:
\[
\cos y + \sin x = 1
\]
**Solution Steps:**
1. **Differentiate the Equation:**
Start by applying implicit differentiation to both sides of the equation with respect to \( x \):
\[
\frac{d}{dx}(\cos y) + \frac{d}{dx}(\sin x) = \frac{d}{dx}(1)
\]
2. **Applying Derivatives:**
\[
-\sin y \cdot \frac{dy}{dx} + \cos x = 0
\]
3. **Solve for the First Derivative \( y' \):**
Re-arrange to solve for \( \frac{dy}{dx} \) (denote \( y' \) as \( \frac{dy}{dx} \)):
\[
y' = \frac{\cos x}{\sin y}
\]
4. **Find the Second Derivative \( y'' \):**
Differentiate \( y' = \frac{\cos x}{\sin y} \) with respect to \( x \):
\[
y'' = \frac{d}{dx} \left( \frac{\cos x}{\sin y} \right)
\]
5. **Use the Quotient Rule:**
The quotient rule is:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]
Here, \( u = \cos x \) and \( v = \sin y \).
\[
\frac{du}{dx} = -\sin x, \quad \frac{dv}{dx} = \cos y \cdot y'
\]
Substituting into the quotient rule:
\[
y'' = \frac{\sin y (-\sin x) - \cos x (\cos y \cdot y')}{(\sin y)^2}
\]
6. **Substitute \( y' = \frac{\cos x}{\sin y} \):**
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