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:**
Find the derivative of the function.
\[ f(x) = \frac{x}{\sqrt{x^3 + 8}} \]
**Solution Overview:**
To find the derivative of this function, we need to apply the quotient rule, which is used to differentiate functions of the form \(\frac{u(x)}{v(x)}\). The formula for the quotient rule is:
\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
In this problem:
- \( u(x) = x \)
- \( v(x) = \sqrt{x^3 + 8} \)
Start by finding \( u'(x) \) and \( v'(x) \):
- \( u'(x) = 1 \)
- \( v(x) = (x^3 + 8)^{1/2} \)
Using the chain rule to differentiate \( v(x) \):
\[ v'(x) = \frac{1}{2}(x^3 + 8)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 8}} \]
Apply the quotient rule:
\[ f'(x) = \frac{1 \cdot \sqrt{x^3 + 8} - x \cdot \frac{3x^2}{2\sqrt{x^3 + 8}}}{(\sqrt{x^3 + 8})^2} \]
Simplify the expression:
The numerator becomes:
\[ \sqrt{x^3 + 8} - \frac{3x^3}{2\sqrt{x^3 + 8}} \]
Thus, you get:
\[ f'(x) = \frac{2(x^3 + 8) - 3x^3}{2(x^3 + 8)\sqrt{x^3 + 8}} \]
Further simplification gives:
\[ f'(x) = \frac{16 - x^3}{2(x^3 + 8)\sqrt{x^3 + 8}} \]
This is the derivative of \( f(x) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd523ae2e-d1e3-48cd-8a9c-04e15052448b%2Fb311fde2-d7e8-49ca-84cd-e2f5063c35d6%2F808m4q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the derivative of the function.
\[ f(x) = \frac{x}{\sqrt{x^3 + 8}} \]
**Solution Overview:**
To find the derivative of this function, we need to apply the quotient rule, which is used to differentiate functions of the form \(\frac{u(x)}{v(x)}\). The formula for the quotient rule is:
\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
In this problem:
- \( u(x) = x \)
- \( v(x) = \sqrt{x^3 + 8} \)
Start by finding \( u'(x) \) and \( v'(x) \):
- \( u'(x) = 1 \)
- \( v(x) = (x^3 + 8)^{1/2} \)
Using the chain rule to differentiate \( v(x) \):
\[ v'(x) = \frac{1}{2}(x^3 + 8)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 8}} \]
Apply the quotient rule:
\[ f'(x) = \frac{1 \cdot \sqrt{x^3 + 8} - x \cdot \frac{3x^2}{2\sqrt{x^3 + 8}}}{(\sqrt{x^3 + 8})^2} \]
Simplify the expression:
The numerator becomes:
\[ \sqrt{x^3 + 8} - \frac{3x^3}{2\sqrt{x^3 + 8}} \]
Thus, you get:
\[ f'(x) = \frac{2(x^3 + 8) - 3x^3}{2(x^3 + 8)\sqrt{x^3 + 8}} \]
Further simplification gives:
\[ f'(x) = \frac{16 - x^3}{2(x^3 + 8)\sqrt{x^3 + 8}} \]
This is the derivative of \( f(x) \).
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