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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![**Problem Statement:**
Find a formula for \( y' \) and determine the slope \( y' \big|_{x=3} \) for the following function.
\[ y = \frac{-8e^x}{-7e^x + 4} \]
**Solution Approach:**
To tackle this problem, we need to follow these steps:
1. Apply the quotient rule to find the derivative \( y' \).
2. Substitute \( x = 3 \) into the derivative formula to find the slope at this specific point.
**1. Derivative of the Function Using the Quotient Rule:**
The quotient rule for differentiating a function of the form \( \frac{u(x)}{v(x)} \) is given by:
\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
In this case:
- \( u(x) = -8e^x \)
- \( v(x) = -7e^x + 4 \)
Using the chain rule, we find the derivatives of \( u \) and \( v \):
- \( u'(x) = -8e^x \) (since the derivative of \( e^x \) is \( e^x \))
- \( v'(x) = -7e^x \) (similarly because the derivative of \( e^x \) is \( e^x \))
Now, applying the quotient rule:
\[ y' = \frac{(-8e^x)(-7e^x + 4) - (-8e^x)(-7e^x)}{(-7e^x + 4)^2} \]
Simplify the numerator:
\[ y' = \frac{56e^{2x} - 32e^x + 56e^{2x}}{(-7e^x + 4)^2} \]
\[ y' = \frac{112e^{2x} - 32e^x}{(-7e^x + 4)^2} \]
**2. Evaluate the Derivative at \( x = 3 \):**
Substitute \( x = 3 \) into the derivative formula to find the slope at this value:
\[ y' \big|_{x=3} = \frac{112e^{6} - 32e^{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F323c4818-e3ec-479b-a7ba-1a514193e117%2F9956204e-2b01-4c10-aa99-2833b0bf881c%2F82e6ovr_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find a formula for \( y' \) and determine the slope \( y' \big|_{x=3} \) for the following function.
\[ y = \frac{-8e^x}{-7e^x + 4} \]
**Solution Approach:**
To tackle this problem, we need to follow these steps:
1. Apply the quotient rule to find the derivative \( y' \).
2. Substitute \( x = 3 \) into the derivative formula to find the slope at this specific point.
**1. Derivative of the Function Using the Quotient Rule:**
The quotient rule for differentiating a function of the form \( \frac{u(x)}{v(x)} \) is given by:
\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
In this case:
- \( u(x) = -8e^x \)
- \( v(x) = -7e^x + 4 \)
Using the chain rule, we find the derivatives of \( u \) and \( v \):
- \( u'(x) = -8e^x \) (since the derivative of \( e^x \) is \( e^x \))
- \( v'(x) = -7e^x \) (similarly because the derivative of \( e^x \) is \( e^x \))
Now, applying the quotient rule:
\[ y' = \frac{(-8e^x)(-7e^x + 4) - (-8e^x)(-7e^x)}{(-7e^x + 4)^2} \]
Simplify the numerator:
\[ y' = \frac{56e^{2x} - 32e^x + 56e^{2x}}{(-7e^x + 4)^2} \]
\[ y' = \frac{112e^{2x} - 32e^x}{(-7e^x + 4)^2} \]
**2. Evaluate the Derivative at \( x = 3 \):**
Substitute \( x = 3 \) into the derivative formula to find the slope at this value:
\[ y' \big|_{x=3} = \frac{112e^{6} - 32e^{
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