Step 4 Finally, determine(-3x + 7) and (x + 7) and simplify the result. dx dx 1. du 1 dv + U dx V dx 1 d -3x + 7/dx 1 = (-3x+7) ([ -3x + 7) + h'(x)=[In[(-3x + 7)(x + 7)]] = dx + 1 X+7 1 X+7 =(x + 7) dx 1 ]) + (x + 7) (¹) -3x + 7 Therefore, if h(x) = In[(-3x + 7)(x + 7)], then we have the following result.
Step 4 Finally, determine(-3x + 7) and (x + 7) and simplify the result. dx dx 1. du 1 dv + U dx V dx 1 d -3x + 7/dx 1 = (-3x+7) ([ -3x + 7) + h'(x)=[In[(-3x + 7)(x + 7)]] = dx + 1 X+7 1 X+7 =(x + 7) dx 1 ]) + (x + 7) (¹) -3x + 7 Therefore, if h(x) = In[(-3x + 7)(x + 7)], then we have the following result.
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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![**Step 4**
Finally, determine \( \frac{d}{dx}(-3x + 7) \) and \( \frac{d}{dx}(x + 7) \) and simplify the result.
\[
\frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} = \left( \frac{1}{-3x + 7} \right) \left( \frac{d}{dx}(-3x + 7) \right) + \left( \frac{1}{x + 7} \right) \left( \frac{d}{dx}(x + 7) \right)
\]
\[
= \left( \frac{1}{-3x + 7} \right)(\underline{\hspace{1cm}}) + \left( \frac{1}{x + 7} \right)(1)
\]
\[
= \frac{\underline{\hspace{0.5cm}}}{-3x + 7} + \frac{1}{x + 7}
\]
Therefore, if \( h(x) = \ln[(-3x + 7)(x + 7)] \), then we have the following result.
\[
h'(x) = \frac{d}{dx} [ \ln[(-3x + 7)(x + 7)] ] = \underline{\hspace{3cm}}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd48171ba-d1cc-47b1-bf9c-8041143384f0%2F1c2317af-3b91-4962-a17a-422121b28079%2Fimdrpw7_processed.png&w=3840&q=75)
Transcribed Image Text:**Step 4**
Finally, determine \( \frac{d}{dx}(-3x + 7) \) and \( \frac{d}{dx}(x + 7) \) and simplify the result.
\[
\frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} = \left( \frac{1}{-3x + 7} \right) \left( \frac{d}{dx}(-3x + 7) \right) + \left( \frac{1}{x + 7} \right) \left( \frac{d}{dx}(x + 7) \right)
\]
\[
= \left( \frac{1}{-3x + 7} \right)(\underline{\hspace{1cm}}) + \left( \frac{1}{x + 7} \right)(1)
\]
\[
= \frac{\underline{\hspace{0.5cm}}}{-3x + 7} + \frac{1}{x + 7}
\]
Therefore, if \( h(x) = \ln[(-3x + 7)(x + 7)] \), then we have the following result.
\[
h'(x) = \frac{d}{dx} [ \ln[(-3x + 7)(x + 7)] ] = \underline{\hspace{3cm}}
\]
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