Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter6: Applications Of The Derivative
Section6.CR: Chapter 6 Review
Problem 48CR
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Question
![### Problem Statement
**2. Use implicit differentiation to find \( y' \) for the curve:**
\[ \cos x \sin y = \frac{1}{4} \]
### Explanation
To solve this problem, you need to use the technique of implicit differentiation. In implicit differentiation, both \( x \) and \( y \) are functions of some variable, often \( t \). The following steps outline the process:
1. **Differentiate both sides of the equation with respect to \( x \)**:
Given:
\[ \cos x \sin y = \frac{1}{4} \]
Differentiate using the product rule on the left side:
\[ \frac{d}{dx} (\cos x \sin y) = \frac{d}{dx} \frac{1}{4} \]
2. **Apply the product rule and chain rule**:
Recall that the product rule states \( \frac{d}{dx} [u v] = u'v + uv' \), and the chain rule is used when differentiating a composite function.
Here:
\[ \frac{d}{dx} (\cos x \sin y) = (\frac{d}{dx} \cos x) \sin y + \cos x \frac{d}{dx} \sin y \]
Using the derivatives:
\[ \frac{d}{dx} (\cos x) = -\sin x \]
\[ \frac{d}{dx} (\sin y) = \cos y \cdot \frac{dy}{dx} = \cos y \cdot y'\]
Substituting these back in:
\[ -\sin x \sin y + \cos x \cos y \frac{dy}{dx} = 0 \]
3. **Solve for \( y' \)**:
\[ \cos x \cos y \frac{dy}{dx} = \sin x \sin y \]
\[ \frac{dy}{dx} = y' = \frac{\sin x \sin y}{\cos x \cos y} \]
\[ y' = \tan x \tan y \]
Thus, the derivative \( y' \) for the given curve using implicit differentiation is:
\[ y' = \tan x \tan y \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52db8ee6-33d1-4aac-bbe4-ebc771de1e49%2F65eed60e-2e05-45bb-9780-71c09633f749%2Froorlm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**2. Use implicit differentiation to find \( y' \) for the curve:**
\[ \cos x \sin y = \frac{1}{4} \]
### Explanation
To solve this problem, you need to use the technique of implicit differentiation. In implicit differentiation, both \( x \) and \( y \) are functions of some variable, often \( t \). The following steps outline the process:
1. **Differentiate both sides of the equation with respect to \( x \)**:
Given:
\[ \cos x \sin y = \frac{1}{4} \]
Differentiate using the product rule on the left side:
\[ \frac{d}{dx} (\cos x \sin y) = \frac{d}{dx} \frac{1}{4} \]
2. **Apply the product rule and chain rule**:
Recall that the product rule states \( \frac{d}{dx} [u v] = u'v + uv' \), and the chain rule is used when differentiating a composite function.
Here:
\[ \frac{d}{dx} (\cos x \sin y) = (\frac{d}{dx} \cos x) \sin y + \cos x \frac{d}{dx} \sin y \]
Using the derivatives:
\[ \frac{d}{dx} (\cos x) = -\sin x \]
\[ \frac{d}{dx} (\sin y) = \cos y \cdot \frac{dy}{dx} = \cos y \cdot y'\]
Substituting these back in:
\[ -\sin x \sin y + \cos x \cos y \frac{dy}{dx} = 0 \]
3. **Solve for \( y' \)**:
\[ \cos x \cos y \frac{dy}{dx} = \sin x \sin y \]
\[ \frac{dy}{dx} = y' = \frac{\sin x \sin y}{\cos x \cos y} \]
\[ y' = \tan x \tan y \]
Thus, the derivative \( y' \) for the given curve using implicit differentiation is:
\[ y' = \tan x \tan y \]
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