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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![**Differentiate**
\[ \frac{d}{dx} \left( \frac{4 - xe^x}{x + e^x} \right) = \quad \underline{\hspace{5cm}} \]
In this problem, we are tasked with differentiating the given function:
\[ \frac{d}{dx} \left( \frac{4 - xe^x}{x + e^x} \right) \]
To solve this, we need to apply the quotient rule of differentiation. The quotient rule states that if we have a function of the form:
\[ \frac{u(x)}{v(x)} \]
where \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative is given by:
\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]
In our case, let:
\[ u(x) = 4 - xe^x \]
\[ v(x) = x + e^x \]
First, we need to find the derivatives \( u'(x) \) and \( v'(x) \):
For \( u(x) = 4 - xe^x \):
\[ u'(x) = \frac{d}{dx}(4) - \frac{d}{dx}(xe^x) \]
\[ u'(x) = 0 - \left( x \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(x) \right) \]
\[ u'(x) = 0 - \left( x \cdot e^x + e^x \cdot 1 \right) \]
\[ u'(x) = -\left( x e^x + e^x \right) \]
\[ u'(x) = -e^x (x + 1) \]
For \( v(x) = x + e^x \):
\[ v'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(e^x) \]
\[ v'(x) = 1 + e^x \]
Now, substitute \( u(x) \), \( u](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F926c3fce-ea0b-442b-8045-2d4039975256%2F18d7ec73-a667-4b9d-ba7c-5a6ab6d82107%2Fq78s8c_processed.png&w=3840&q=75)
Transcribed Image Text:**Differentiate**
\[ \frac{d}{dx} \left( \frac{4 - xe^x}{x + e^x} \right) = \quad \underline{\hspace{5cm}} \]
In this problem, we are tasked with differentiating the given function:
\[ \frac{d}{dx} \left( \frac{4 - xe^x}{x + e^x} \right) \]
To solve this, we need to apply the quotient rule of differentiation. The quotient rule states that if we have a function of the form:
\[ \frac{u(x)}{v(x)} \]
where \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative is given by:
\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]
In our case, let:
\[ u(x) = 4 - xe^x \]
\[ v(x) = x + e^x \]
First, we need to find the derivatives \( u'(x) \) and \( v'(x) \):
For \( u(x) = 4 - xe^x \):
\[ u'(x) = \frac{d}{dx}(4) - \frac{d}{dx}(xe^x) \]
\[ u'(x) = 0 - \left( x \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(x) \right) \]
\[ u'(x) = 0 - \left( x \cdot e^x + e^x \cdot 1 \right) \]
\[ u'(x) = -\left( x e^x + e^x \right) \]
\[ u'(x) = -e^x (x + 1) \]
For \( v(x) = x + e^x \):
\[ v'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(e^x) \]
\[ v'(x) = 1 + e^x \]
Now, substitute \( u(x) \), \( u
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