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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![**Problem Statement:**
Find \( k'(x) \) if
\[ k(x) = e^{\left(-5x^{5/3} - \frac{1}{4}x^{-1/5}\right)}. \]
**Instructions:**
Provide your answer below:
\[ k'(x) = \Box \]
---
**Solution Guide:**
The problem requires finding the derivative of the function \( k(x) \), which is expressed in exponential form. This problem involves using rules of calculus, particularly focusing on the chain rule, due to the composition of functions.
**Detailed Explanation:**
The derivative of an exponential function of the form \( e^{f(x)} \) can be obtained using the chain rule, which states: the derivative of \( e^{f(x)} \) is \( e^{f(x)} \cdot f'(x) \). Therefore, you must first find \( f'(x) \), where:
\[ f(x) = -5x^{5/3} - \frac{1}{4}x^{-1/5} \]
To find \( f'(x) \), differentiate each term:
1. The derivative of the first term \(-5x^{5/3}\) is \(-\frac{5 \cdot 5}{3}x^{5/3 - 1} = -\frac{25}{3}x^{2/3}\).
2. The derivative of the second term \(-\frac{1}{4}x^{-1/5}\) is \(\left(-\frac{1}{4} \cdot -\frac{1}{5}\right)x^{-1/5 - 1} = \frac{1}{20}x^{-6/5}\).
Combine these to get:
\[ f'(x) = -\frac{25}{3}x^{2/3} + \frac{1}{20}x^{-6/5} \]
Using the chain rule, the derivative \( k'(x) \) is:
\[ k'(x) = e^{\left(-5x^{5/3} - \frac{1}{4}x^{-1/5}\right)} \left(-\frac{25}{3}x^{2/3} + \frac{1}{20}x^{-6/5}\right) \]
Place this expression in the blank box](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad396f38-0950-4136-90e0-b177d8f23e70%2F3d32af53-4f2d-4aab-915e-04033d41a1a0%2F3fzc3r_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \( k'(x) \) if
\[ k(x) = e^{\left(-5x^{5/3} - \frac{1}{4}x^{-1/5}\right)}. \]
**Instructions:**
Provide your answer below:
\[ k'(x) = \Box \]
---
**Solution Guide:**
The problem requires finding the derivative of the function \( k(x) \), which is expressed in exponential form. This problem involves using rules of calculus, particularly focusing on the chain rule, due to the composition of functions.
**Detailed Explanation:**
The derivative of an exponential function of the form \( e^{f(x)} \) can be obtained using the chain rule, which states: the derivative of \( e^{f(x)} \) is \( e^{f(x)} \cdot f'(x) \). Therefore, you must first find \( f'(x) \), where:
\[ f(x) = -5x^{5/3} - \frac{1}{4}x^{-1/5} \]
To find \( f'(x) \), differentiate each term:
1. The derivative of the first term \(-5x^{5/3}\) is \(-\frac{5 \cdot 5}{3}x^{5/3 - 1} = -\frac{25}{3}x^{2/3}\).
2. The derivative of the second term \(-\frac{1}{4}x^{-1/5}\) is \(\left(-\frac{1}{4} \cdot -\frac{1}{5}\right)x^{-1/5 - 1} = \frac{1}{20}x^{-6/5}\).
Combine these to get:
\[ f'(x) = -\frac{25}{3}x^{2/3} + \frac{1}{20}x^{-6/5} \]
Using the chain rule, the derivative \( k'(x) \) is:
\[ k'(x) = e^{\left(-5x^{5/3} - \frac{1}{4}x^{-1/5}\right)} \left(-\frac{25}{3}x^{2/3} + \frac{1}{20}x^{-6/5}\right) \]
Place this expression in the blank box
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