e*+ cos x (d) h(x) = tan x (e) y = Vx sec x e* =dx.secX.さャx.d secxrが4リxのsecX.àビ .5ecX-ド4リズ·secx+tanX.どズ·sとx-ビ -4• secx-et x2.secxtan x +x2 secx•ex %3D 2x2
e*+ cos x (d) h(x) = tan x (e) y = Vx sec x e* =dx.secX.さャx.d secxrが4リxのsecX.àビ .5ecX-ド4リズ·secx+tanX.どズ·sとx-ビ -4• secx-et x2.secxtan x +x2 secx•ex %3D 2x2
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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![### Calculus Derivatives
#### Problem (d)
Find the derivative of the function:
\[ h(x) = \frac{e^x + \cos x}{\tan x} \]
#### Problem (e)
Find the derivative of the function:
\[ y = \sqrt{x} \cdot \sec x \cdot e^x \]
**Solution:**
\[
\begin{aligned}
& y = \sqrt{x} \cdot \sec x \cdot e^x \\
& = \frac{d}{dx} \left( x^{1/2} \cdot \sec x \cdot e^x \right) \\
& = \frac{d}{dx} (x^{1/2}) \cdot \sec x \cdot e^x + x^{1/2} \cdot \frac{d}{dx} (\sec x \cdot e^x) \\
& = \frac{1}{2} x^{-1/2} \cdot \sec x \cdot e^x + x^{1/2} \cdot \left( \sec x \cdot \frac{d}{dx} (e^x) + e^x \cdot \frac{d}{dx} (\sec x) \right) \\
& = \frac{1}{2} x^{-1/2} \cdot \sec x \cdot e^x + x^{1/2} \left( \sec x \cdot e^x + e^x \cdot \sec x \cdot \tan x \right) \\
& = \frac{1}{2} x^{-1/2} \cdot \sec x \cdot e^x + x^{1/2} \cdot \sec x \cdot e^x (1 + \tan x) \\
& = \frac{e^x \cdot \sec x}{2 \sqrt{x}} + e^x \cdot \sec x \left( \frac{x^{1/2}}{x^{1/2}} + x^{1/2} \cdot \tan x \right) \\
& = \frac{e^x \cdot \sec x}{2 \sqrt{x}} + e^x \cdot \sec x \left( \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F40fcff51-67cc-4242-b56b-596269d65d1c%2Fbf3e4b23-5cea-43dc-be93-8017ba525f3a%2Fs361uof_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Derivatives
#### Problem (d)
Find the derivative of the function:
\[ h(x) = \frac{e^x + \cos x}{\tan x} \]
#### Problem (e)
Find the derivative of the function:
\[ y = \sqrt{x} \cdot \sec x \cdot e^x \]
**Solution:**
\[
\begin{aligned}
& y = \sqrt{x} \cdot \sec x \cdot e^x \\
& = \frac{d}{dx} \left( x^{1/2} \cdot \sec x \cdot e^x \right) \\
& = \frac{d}{dx} (x^{1/2}) \cdot \sec x \cdot e^x + x^{1/2} \cdot \frac{d}{dx} (\sec x \cdot e^x) \\
& = \frac{1}{2} x^{-1/2} \cdot \sec x \cdot e^x + x^{1/2} \cdot \left( \sec x \cdot \frac{d}{dx} (e^x) + e^x \cdot \frac{d}{dx} (\sec x) \right) \\
& = \frac{1}{2} x^{-1/2} \cdot \sec x \cdot e^x + x^{1/2} \left( \sec x \cdot e^x + e^x \cdot \sec x \cdot \tan x \right) \\
& = \frac{1}{2} x^{-1/2} \cdot \sec x \cdot e^x + x^{1/2} \cdot \sec x \cdot e^x (1 + \tan x) \\
& = \frac{e^x \cdot \sec x}{2 \sqrt{x}} + e^x \cdot \sec x \left( \frac{x^{1/2}}{x^{1/2}} + x^{1/2} \cdot \tan x \right) \\
& = \frac{e^x \cdot \sec x}{2 \sqrt{x}} + e^x \cdot \sec x \left( \
![### Limits in Calculus:
#### Problem (b)
Evaluate:
\[ \lim_{x \to \infty} \frac{5x^2 - 8x^4}{3x^4 + 9} \]
This limit involves a rational function where the degrees of the polynomials in the numerator and the denominator are different.
#### Solution Steps:
1. **Identify the highest degree terms** in both the numerator and the denominator:
\[ \text{Numerator: } -8x^4 \]
\[ \text{Denominator: } 3x^4 \]
2. **Divide both the numerator and the denominator** by \(x^4\), the highest power in the denominator:
\[ \frac{5x^2 - 8x^4}{3x^4 + 9} = \frac{\frac{5x^2}{x^4} - \frac{8x^4}{x^4}}{\frac{3x^4}{x^4} + \frac{9}{x^4}} = \frac{\frac{5}{x^2} - 8}{3 + \frac{9}{x^4}} \]
3. **Take the limit** as \(x\) approaches infinity:
\[ \lim_{x \to \infty} \frac{\frac{5}{x^2} - 8}{3 + \frac{9}{x^4}} \]
As \(x \to \infty\), \(\frac{5}{x^2} \to 0\) and \(\frac{9}{x^4} \to 0\):
\[ \frac{0 - 8}{3 + 0} = \frac{-8}{3} \]
4. **Conclusion**:
\[ \lim_{x \to \infty} \frac{5x^2 - 8x^4}{3x^4 + 9} = -\frac{8}{3} \]
---
#### Problem (c)
Evaluate:
\[ \lim_{x \to 0} \frac{\tan 3x}{5x} \]
This limit involves the tangent function, and the presented form involves \(x\) approaching zero.
#### Solution Steps:
1. **Use the small-angle approximation**:
For small angles](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F40fcff51-67cc-4242-b56b-596269d65d1c%2Fbf3e4b23-5cea-43dc-be93-8017ba525f3a%2Fbjsfb9vh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Limits in Calculus:
#### Problem (b)
Evaluate:
\[ \lim_{x \to \infty} \frac{5x^2 - 8x^4}{3x^4 + 9} \]
This limit involves a rational function where the degrees of the polynomials in the numerator and the denominator are different.
#### Solution Steps:
1. **Identify the highest degree terms** in both the numerator and the denominator:
\[ \text{Numerator: } -8x^4 \]
\[ \text{Denominator: } 3x^4 \]
2. **Divide both the numerator and the denominator** by \(x^4\), the highest power in the denominator:
\[ \frac{5x^2 - 8x^4}{3x^4 + 9} = \frac{\frac{5x^2}{x^4} - \frac{8x^4}{x^4}}{\frac{3x^4}{x^4} + \frac{9}{x^4}} = \frac{\frac{5}{x^2} - 8}{3 + \frac{9}{x^4}} \]
3. **Take the limit** as \(x\) approaches infinity:
\[ \lim_{x \to \infty} \frac{\frac{5}{x^2} - 8}{3 + \frac{9}{x^4}} \]
As \(x \to \infty\), \(\frac{5}{x^2} \to 0\) and \(\frac{9}{x^4} \to 0\):
\[ \frac{0 - 8}{3 + 0} = \frac{-8}{3} \]
4. **Conclusion**:
\[ \lim_{x \to \infty} \frac{5x^2 - 8x^4}{3x^4 + 9} = -\frac{8}{3} \]
---
#### Problem (c)
Evaluate:
\[ \lim_{x \to 0} \frac{\tan 3x}{5x} \]
This limit involves the tangent function, and the presented form involves \(x\) approaching zero.
#### Solution Steps:
1. **Use the small-angle approximation**:
For small angles
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