Step 3 Now write the derivative in terms of x by recalling that u = -3x + 7 and v = x + 7. du 1 += dx V [In(u)] + dx [In(v)] = dx 1 u 1 -3x + 7 d dx dv dx 1 d ] ) + ( x + 7) + (x + 7) X+7 dx

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 3

Now write the derivative in terms of \( x \) by recalling that \( u = -3x + 7 \) and \( v = x + 7 \).

\[
\frac{d}{dx} [\ln(u)] + \frac{d}{dx} [\ln(v)] = \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx}
\]

\[
= \left( \frac{1}{-3x + 7} \right) \frac{d}{dx} \left(\underline{\hspace{1cm}}\right) + \left( \frac{1}{x + 7} \right) \frac{d}{dx} (x + 7)
\] 

In the equations, \( u \) and \( v \) are defined as linear expressions of \( x \). The derivatives of the natural logarithmic functions of \( u \) and \( v \) are found by using the chain rule, where the derivative is split into components of the derivative of the natural log function and the derivative of \( u \) and \( v \) themselves. The terms are then substituted back into the original expression to complete the derivative in terms of \( x \).
Transcribed Image Text:Step 3 Now write the derivative in terms of \( x \) by recalling that \( u = -3x + 7 \) and \( v = x + 7 \). \[ \frac{d}{dx} [\ln(u)] + \frac{d}{dx} [\ln(v)] = \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} \] \[ = \left( \frac{1}{-3x + 7} \right) \frac{d}{dx} \left(\underline{\hspace{1cm}}\right) + \left( \frac{1}{x + 7} \right) \frac{d}{dx} (x + 7) \] In the equations, \( u \) and \( v \) are defined as linear expressions of \( x \). The derivatives of the natural logarithmic functions of \( u \) and \( v \) are found by using the chain rule, where the derivative is split into components of the derivative of the natural log function and the derivative of \( u \) and \( v \) themselves. The terms are then substituted back into the original expression to complete the derivative in terms of \( x \).
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