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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![## Calculus Problem: Differentiation
**Problem 8**: Find \( y' \) if \( y = x^3 + \sin(x) \).
### Solution:
To find the derivative \( y' \) of the function \( y = x^3 + \sin(x) \), we apply the rules of differentiation.
1. Differentiate \( x^3 \) with respect to \( x \):
\[
\frac{d}{dx}(x^3) = 3x^2
\]
2. Differentiate \( \sin(x) \) with respect to \( x \):
\[
\frac{d}{dx}(\sin(x)) = \cos(x)
\]
Combining these results, the derivative \( y' \) is:
\[
y' = 3x^2 + \cos(x)
\]
Thus, the derivative of \( y = x^3 + \sin(x) \) is \( y' = 3x^2 + \cos(x) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F22969543-4b1b-47a8-98d2-f1e5e1e74392%2Fdee695aa-1fc5-4e00-945b-6a623139518a%2Fuc6rntu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Calculus Problem: Differentiation
**Problem 8**: Find \( y' \) if \( y = x^3 + \sin(x) \).
### Solution:
To find the derivative \( y' \) of the function \( y = x^3 + \sin(x) \), we apply the rules of differentiation.
1. Differentiate \( x^3 \) with respect to \( x \):
\[
\frac{d}{dx}(x^3) = 3x^2
\]
2. Differentiate \( \sin(x) \) with respect to \( x \):
\[
\frac{d}{dx}(\sin(x)) = \cos(x)
\]
Combining these results, the derivative \( y' \) is:
\[
y' = 3x^2 + \cos(x)
\]
Thus, the derivative of \( y = x^3 + \sin(x) \) is \( y' = 3x^2 + \cos(x) \).
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