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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Question
![**Problem Statement:**
Let \( h(u) = 4u^3 - \frac{1}{u^2} \).
**Task:**
Find \( h'(u) \). Show your work.
---
**Solution:**
To find \( h'(u) \), we will differentiate each term of the function \( h(u) \) separately using the rules of differentiation.
1. Differentiate the first term \( 4u^3 \).
- The derivative of \( u^n \) is \( nu^{n-1} \).
- Therefore, the derivative of \( 4u^3 \) is:
\[
4 \times 3u^{3-1} = 12u^2
\]
2. Differentiate the second term \( -\frac{1}{u^2} \).
- Rewrite \( \frac{1}{u^2} \) as \( u^{-2} \).
- The derivative of \( u^{-n} \) is \( -nu^{-n-1} \).
- Therefore, the derivative of \( u^{-2} \) is:
\[
-(-2)u^{-2-1} = 2u^{-3}
\]
- This simplifies to \( \frac{2}{u^3} \).
**Final Derivative:**
Combine the results to find \( h'(u) \):
\[
h'(u) = 12u^2 + \frac{2}{u^3}
\]
Thus, the derivative of \( h(u) \) is \( h'(u) = 12u^2 + \frac{2}{u^3} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd556e803-222c-475e-bc6a-44d8250d1cd2%2F7169f523-08bb-45c0-a020-aff1cb32076c%2Fm5n6ixt_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( h(u) = 4u^3 - \frac{1}{u^2} \).
**Task:**
Find \( h'(u) \). Show your work.
---
**Solution:**
To find \( h'(u) \), we will differentiate each term of the function \( h(u) \) separately using the rules of differentiation.
1. Differentiate the first term \( 4u^3 \).
- The derivative of \( u^n \) is \( nu^{n-1} \).
- Therefore, the derivative of \( 4u^3 \) is:
\[
4 \times 3u^{3-1} = 12u^2
\]
2. Differentiate the second term \( -\frac{1}{u^2} \).
- Rewrite \( \frac{1}{u^2} \) as \( u^{-2} \).
- The derivative of \( u^{-n} \) is \( -nu^{-n-1} \).
- Therefore, the derivative of \( u^{-2} \) is:
\[
-(-2)u^{-2-1} = 2u^{-3}
\]
- This simplifies to \( \frac{2}{u^3} \).
**Final Derivative:**
Combine the results to find \( h'(u) \):
\[
h'(u) = 12u^2 + \frac{2}{u^3}
\]
Thus, the derivative of \( h(u) \) is \( h'(u) = 12u^2 + \frac{2}{u^3} \).
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