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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23 and 31
![### Section 12.7: Higher-Order Derivatives
---
**Example Derivation**
Given the equation:
\[ 2y = e^{x/y} \]
Differentiate both sides:
\[ \frac{d}{dx}\left(2y\right) = \frac{d}{dx}\left(e^{x/y}\right) \]
On the left:
\[ 2 \cdot \frac{dy}{dx} \]
On the right, using chain rule:
\[ \frac{e^{x/y}}{y^2} (y - x \cdot \frac{dy}{dx}) \]
**Since** \(y^3 = e^x\) (the original equation):
\[ y^2 \cdot \frac{dy}{dx} = e^x \]
**Solving for** \(\frac{dy}{dx}\):
\[ \Rightarrow \frac{dy}{dx} = \frac{2y - \frac{2}{y}}{2y^2} \]
Final expression without \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{2 \left(\frac{2}{y}\right)}{(2 - y^2)^2} \]
**Now Work Problem 31**
---
### PROBLEMS 12.7
**In Problems 1-20, find the indicated derivatives.**
1. \( y = 4x^3 - 12x^2 + (x + 2)\, y'\)
2. \( y = x + \frac{e^x}{2} + x^{-2}, y^{\prime\prime\prime} \)
3. \( y = e^{5x} + 3y^\prime \)
4. \( y = x^3 \ln(x), y^{\prime\prime\prime} \)
5. \( y = x^{3/2} + e^{3x} \)
6. \( F(g) = \ln(q + t), \frac{dF}{dg} \)
7. \( f(x) = 3 \ln x,\, \, y_{\min} \)
8. \( y = \frac{1}{x^m} \)
9. \( f(q) = \sqrt[3]{2q^7} \)
10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0e2d6f28-e3d7-414c-b1d4-a752131175c2%2F4b865af5-625b-4fde-8424-6e1abc73f597%2F85ihsla.jpeg&w=3840&q=75)
Transcribed Image Text:### Section 12.7: Higher-Order Derivatives
---
**Example Derivation**
Given the equation:
\[ 2y = e^{x/y} \]
Differentiate both sides:
\[ \frac{d}{dx}\left(2y\right) = \frac{d}{dx}\left(e^{x/y}\right) \]
On the left:
\[ 2 \cdot \frac{dy}{dx} \]
On the right, using chain rule:
\[ \frac{e^{x/y}}{y^2} (y - x \cdot \frac{dy}{dx}) \]
**Since** \(y^3 = e^x\) (the original equation):
\[ y^2 \cdot \frac{dy}{dx} = e^x \]
**Solving for** \(\frac{dy}{dx}\):
\[ \Rightarrow \frac{dy}{dx} = \frac{2y - \frac{2}{y}}{2y^2} \]
Final expression without \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{2 \left(\frac{2}{y}\right)}{(2 - y^2)^2} \]
**Now Work Problem 31**
---
### PROBLEMS 12.7
**In Problems 1-20, find the indicated derivatives.**
1. \( y = 4x^3 - 12x^2 + (x + 2)\, y'\)
2. \( y = x + \frac{e^x}{2} + x^{-2}, y^{\prime\prime\prime} \)
3. \( y = e^{5x} + 3y^\prime \)
4. \( y = x^3 \ln(x), y^{\prime\prime\prime} \)
5. \( y = x^{3/2} + e^{3x} \)
6. \( F(g) = \ln(q + t), \frac{dF}{dg} \)
7. \( f(x) = 3 \ln x,\, \, y_{\min} \)
8. \( y = \frac{1}{x^m} \)
9. \( f(q) = \sqrt[3]{2q^7} \)
10
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