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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![**Find the derivative of the function:**
\[ f(x) = \arctan(3x)\]
The derivative of this function can be found using the chain rule.
\[ f(x) = \arctan(u) \] where \[ u = 3x \]
The derivative of \(\arctan(u)\) is \(\dfrac{1}{1+u^2}\). Therefore, we need to multiply this by the derivative of \(u\).
\[ f'(x) = \dfrac{d}{dx} \left(\arctan(3x) \right) \]
Using the chain rule:
\[ f'(x) = \dfrac{1}{1+(3x)^2} \cdot \dfrac{d}{dx} (3x) \]
The derivative of \(3x\) is 3, so we have:
\[ f'(x) = \dfrac{3}{1+9x^2} \]
Thus, the derivative of the function \( f(x) = \arctan(3x) \) is:
\[ f'(x) = \dfrac{3}{9x^2 + 1} \]
This derivative is displayed in the image as a simple formula input:
\[ \frac{3}{9x^2 + 1} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffef832d7-febb-470a-ad68-b786ac1e067f%2F8bd2d198-27f7-488b-bb67-1697307731c7%2Fa93smbp_processed.png&w=3840&q=75)
Transcribed Image Text:**Find the derivative of the function:**
\[ f(x) = \arctan(3x)\]
The derivative of this function can be found using the chain rule.
\[ f(x) = \arctan(u) \] where \[ u = 3x \]
The derivative of \(\arctan(u)\) is \(\dfrac{1}{1+u^2}\). Therefore, we need to multiply this by the derivative of \(u\).
\[ f'(x) = \dfrac{d}{dx} \left(\arctan(3x) \right) \]
Using the chain rule:
\[ f'(x) = \dfrac{1}{1+(3x)^2} \cdot \dfrac{d}{dx} (3x) \]
The derivative of \(3x\) is 3, so we have:
\[ f'(x) = \dfrac{3}{1+9x^2} \]
Thus, the derivative of the function \( f(x) = \arctan(3x) \) is:
\[ f'(x) = \dfrac{3}{9x^2 + 1} \]
This derivative is displayed in the image as a simple formula input:
\[ \frac{3}{9x^2 + 1} \]
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