Find the derivative of the function. f(x) = X + 8

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...
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 \( f(x) \).
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 \( f(x) \).
Expert Solution
Step 1

Here, we need to find the derivative of f(x)=xx3+8.

So,

         f'(x)=xx3+8'ApplytheQuotientRule:  fg'=f'·g-g'·fg2.      =x'x3+8-x3+8'xx3+82      =1·x3+8-3x22x3+8xx3+82      =-x3+162x3+8x3+8

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