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.**
Given function:
\[ f(z) = e^{\arcsin(z^5)} \]
Find the derivative:
\[ f'(z) = \]
(Note: A red 'X' indicates the solution is incorrect or incomplete.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdce9ba03-80ad-4ef0-9d01-d879c68da607%2F755e879d-2cb8-496a-ba46-e74cf6a6d8fd%2Ftgqoiqi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find the derivative of the function.**
Given function:
\[ f(z) = e^{\arcsin(z^5)} \]
Find the derivative:
\[ f'(z) = \]
(Note: A red 'X' indicates the solution is incorrect or incomplete.)
![**Problem Statement:**
Find the derivative of the function.
\[ y = (\tan^{-1}(3x))^2 \]
**Solution:**
To find the derivative \( y' \), apply the chain rule and the derivative of the inverse tangent function.
1. Let \( u = \tan^{-1}(3x) \), so \( y = u^2 \).
2. The derivative of \( u^2 \) with respect to \( u \) is \( 2u \).
3. Now, find the derivative of \( u = \tan^{-1}(3x) \) with respect to \( x \).
The derivative of \(\tan^{-1}(3x)\) is:
\[
\frac{d}{dx}[\tan^{-1}(3x)] = \frac{1}{1 + (3x)^2} \cdot 3 = \frac{3}{1 + 9x^2}
\]
4. Using the chain rule:
\[
y' = 2u \cdot \frac{du}{dx} = 2(\tan^{-1}(3x)) \cdot \frac{3}{1 + 9x^2}
\]
5. Simplify:
\[
y' = \frac{6\tan^{-1}(3x)}{1 + 9x^2}
\]
Thus, the derivative \( y' \) is:
\[
y' = \frac{6\tan^{-1}(3x)}{1 + 9x^2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdce9ba03-80ad-4ef0-9d01-d879c68da607%2F755e879d-2cb8-496a-ba46-e74cf6a6d8fd%2F7ij99zk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the derivative of the function.
\[ y = (\tan^{-1}(3x))^2 \]
**Solution:**
To find the derivative \( y' \), apply the chain rule and the derivative of the inverse tangent function.
1. Let \( u = \tan^{-1}(3x) \), so \( y = u^2 \).
2. The derivative of \( u^2 \) with respect to \( u \) is \( 2u \).
3. Now, find the derivative of \( u = \tan^{-1}(3x) \) with respect to \( x \).
The derivative of \(\tan^{-1}(3x)\) is:
\[
\frac{d}{dx}[\tan^{-1}(3x)] = \frac{1}{1 + (3x)^2} \cdot 3 = \frac{3}{1 + 9x^2}
\]
4. Using the chain rule:
\[
y' = 2u \cdot \frac{du}{dx} = 2(\tan^{-1}(3x)) \cdot \frac{3}{1 + 9x^2}
\]
5. Simplify:
\[
y' = \frac{6\tan^{-1}(3x)}{1 + 9x^2}
\]
Thus, the derivative \( y' \) is:
\[
y' = \frac{6\tan^{-1}(3x)}{1 + 9x^2}
\]
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