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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![### Calculating the Derivative of a Function
**Question:**
What is \( f'(x) \) if \( f(x) = \frac{40x^5}{e^x} \)?
**Explanation:**
To find the derivative \( f'(x) \), utilize the quotient rule, which is applicable to functions of the form \( \frac{u(x)}{v(x)} \). The rule states:
\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
Here, \( u(x) = 40x^5 \) and \( v(x) = e^x \).
**Steps:**
1. **Differentiate \( u(x) = 40x^5 \):**
- \( u'(x) = 200x^4 \).
2. **Differentiate \( v(x) = e^x \):**
- \( v'(x) = e^x \).
3. **Apply the Quotient Rule:**
- Plug the derivatives into the quotient rule formula:
\[
f'(x) = \frac{(200x^4) \cdot (e^x) - (40x^5) \cdot (e^x)}{(e^x)^2}
\]
4. **Simplify:**
- Combine and factor the numerator:
\[
f'(x) = \frac{e^x(200x^4 - 40x^5)}{e^{2x}}
\]
- Simplify further:
\[
f'(x) = \frac{e^x \cdot 40x^4(5 - x)}{e^{2x}}
\]
5. **Final Expression:**
- Cancel \( e^x \):
\[
f'(x) = \frac{40x^4(5 - x)}{e^x}
\]
Thus, the derivative \( f'(x) = \frac{40x^4(5 - x)}{e^x} \).
This solution showcases the application of the quotient rule and simplification steps in derivative calculation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbce27ae5-8d1b-4482-9174-f12c6de24a40%2F781b67c8-da07-4a45-817b-cff416278a1c%2Fc6ujzp_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Derivative of a Function
**Question:**
What is \( f'(x) \) if \( f(x) = \frac{40x^5}{e^x} \)?
**Explanation:**
To find the derivative \( f'(x) \), utilize the quotient rule, which is applicable to functions of the form \( \frac{u(x)}{v(x)} \). The rule states:
\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
Here, \( u(x) = 40x^5 \) and \( v(x) = e^x \).
**Steps:**
1. **Differentiate \( u(x) = 40x^5 \):**
- \( u'(x) = 200x^4 \).
2. **Differentiate \( v(x) = e^x \):**
- \( v'(x) = e^x \).
3. **Apply the Quotient Rule:**
- Plug the derivatives into the quotient rule formula:
\[
f'(x) = \frac{(200x^4) \cdot (e^x) - (40x^5) \cdot (e^x)}{(e^x)^2}
\]
4. **Simplify:**
- Combine and factor the numerator:
\[
f'(x) = \frac{e^x(200x^4 - 40x^5)}{e^{2x}}
\]
- Simplify further:
\[
f'(x) = \frac{e^x \cdot 40x^4(5 - x)}{e^{2x}}
\]
5. **Final Expression:**
- Cancel \( e^x \):
\[
f'(x) = \frac{40x^4(5 - x)}{e^x}
\]
Thus, the derivative \( f'(x) = \frac{40x^4(5 - x)}{e^x} \).
This solution showcases the application of the quotient rule and simplification steps in derivative calculation.
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