d tant lea) de tan' L) Itx? 2x THlege THe" ux 2) la 2x ln x Ids 2 lox t 2 y de 4(2/ax+2) 3) fL7)-6 gi(7)=4 (x=7) t f(G) し -50 t 24 -26
d tant lea) de tan' L) Itx? 2x THlege THe" ux 2) la 2x ln x Ids 2 lox t 2 y de 4(2/ax+2) 3) fL7)-6 gi(7)=4 (x=7) t f(G) し -50 t 24 -26
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Calculus Homework Solutions
The following notes cover derivative problem solutions typically encountered in a Calculus course.
---
#### Problem 1
Evaluate the derivative of the function \( \tan^{-1}(e^x) \):
\[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] \]
We know that:
\[ \frac{d}{dx} \left[ \tan^{-1}(x) \right] = \frac{1}{1 + x^2} \]
Using the chain rule:
\[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + (e^x)^2} \cdot \frac{d}{dx}(e^x) \]
Simplify the expression:
\[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + e^{2x}} \cdot e^x = \frac{e^x}{1 + e^{2x}} = \frac{2e^{2x}}{1 + e^{2x}} \]
---
#### Problem 2
Given the function \( y = 2x \ln x \), find \( \frac{dy}{dx} \):
Use the product rule:
\[ y = 2x \ln x \]
\[ \frac{dy}{dx} = 2 (\ln x) + 2x \left( \frac{1}{x} \right) \]
Simplify the expression:
\[ \frac{dy}{dx} = 2 (\ln x) + 2 \]
Express \( \frac{dy}{dx} \) in terms of \( y \):
\[ \frac{dy}{dx} = y (2 \ln x + 2) \]
---
#### Problem 3
Given the functions and their derivatives:
\[ f(7) = 6, \quad f'(7) = 10 \]
\[ g(7) = -5, \quad g'(7) = 4 \]
Find \( h'(x) \) for \( h(x) = f(x) g(x) \) at \( x = 7 \):
Using the product rule:
\[ h'(x) = f'(x) g(x) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7369c9e1-cd27-40fa-b19e-4e793313e403%2F8f0b906d-15ef-4974-bab6-893c807c2bda%2Fehnmel6.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Homework Solutions
The following notes cover derivative problem solutions typically encountered in a Calculus course.
---
#### Problem 1
Evaluate the derivative of the function \( \tan^{-1}(e^x) \):
\[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] \]
We know that:
\[ \frac{d}{dx} \left[ \tan^{-1}(x) \right] = \frac{1}{1 + x^2} \]
Using the chain rule:
\[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + (e^x)^2} \cdot \frac{d}{dx}(e^x) \]
Simplify the expression:
\[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + e^{2x}} \cdot e^x = \frac{e^x}{1 + e^{2x}} = \frac{2e^{2x}}{1 + e^{2x}} \]
---
#### Problem 2
Given the function \( y = 2x \ln x \), find \( \frac{dy}{dx} \):
Use the product rule:
\[ y = 2x \ln x \]
\[ \frac{dy}{dx} = 2 (\ln x) + 2x \left( \frac{1}{x} \right) \]
Simplify the expression:
\[ \frac{dy}{dx} = 2 (\ln x) + 2 \]
Express \( \frac{dy}{dx} \) in terms of \( y \):
\[ \frac{dy}{dx} = y (2 \ln x + 2) \]
---
#### Problem 3
Given the functions and their derivatives:
\[ f(7) = 6, \quad f'(7) = 10 \]
\[ g(7) = -5, \quad g'(7) = 4 \]
Find \( h'(x) \) for \( h(x) = f(x) g(x) \) at \( x = 7 \):
Using the product rule:
\[ h'(x) = f'(x) g(x) +
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