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 \(\frac{dy}{dt}\).
Given:
\[ y = 3t(2t^2 - 5)^5 \]
**Solution Steps**
1. **Differentiate \( y \) with respect to \( t \).**
You are given the function \( y = 3t(2t^2 - 5)^5 \). To find \(\frac{dy}{dt}\), apply the product rule and the chain rule.
2. **Using the Product Rule:**
The product rule states that if you have a function \( y = u(t) \cdot v(t) \), then the derivative \( \frac{dy}{dt} = u'(t)v(t) + u(t)v'(t) \).
- Here, let:
- \( u(t) = 3t \)
- \( v(t) = (2t^2 - 5)^5 \)
3. **Differentiate \( u(t) \) and \( v(t) \):**
- \( \frac{du}{dt} = 3 \)
- For \( v(t) \), apply the chain rule:
\[
\frac{dv}{dt} = 5(2t^2 - 5)^4 \cdot \frac{d}{dt}(2t^2 - 5)
\]
- Differentiate the inside function \( 2t^2 - 5 \):
\[
\frac{d}{dt}(2t^2 - 5) = 4t
\]
- Therefore,
\[
\frac{dv}{dt} = 5(2t^2 - 5)^4 \times 4t = 20t(2t^2 - 5)^4
\]
4. **Substitute in the Product Rule:**
\[
\frac{dy}{dt} = 3(2t^2 - 5)^5 + 3t \cdot 20t(2t^2 - 5)^4
\]
5. **Simplify the Expression:**
\[
\frac{dy}{dt} = 3(2t^2 - 5)^5 + 60t^2(2t^2 - 5)^4
\]
This expression gives the rate of change of \( y \) with](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc7deb2c-0862-415e-88b2-023e0d274874%2Ff232ffc1-f482-44c3-a702-a95d3c047730%2Fnzmi61_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find \(\frac{dy}{dt}\).
Given:
\[ y = 3t(2t^2 - 5)^5 \]
**Solution Steps**
1. **Differentiate \( y \) with respect to \( t \).**
You are given the function \( y = 3t(2t^2 - 5)^5 \). To find \(\frac{dy}{dt}\), apply the product rule and the chain rule.
2. **Using the Product Rule:**
The product rule states that if you have a function \( y = u(t) \cdot v(t) \), then the derivative \( \frac{dy}{dt} = u'(t)v(t) + u(t)v'(t) \).
- Here, let:
- \( u(t) = 3t \)
- \( v(t) = (2t^2 - 5)^5 \)
3. **Differentiate \( u(t) \) and \( v(t) \):**
- \( \frac{du}{dt} = 3 \)
- For \( v(t) \), apply the chain rule:
\[
\frac{dv}{dt} = 5(2t^2 - 5)^4 \cdot \frac{d}{dt}(2t^2 - 5)
\]
- Differentiate the inside function \( 2t^2 - 5 \):
\[
\frac{d}{dt}(2t^2 - 5) = 4t
\]
- Therefore,
\[
\frac{dv}{dt} = 5(2t^2 - 5)^4 \times 4t = 20t(2t^2 - 5)^4
\]
4. **Substitute in the Product Rule:**
\[
\frac{dy}{dt} = 3(2t^2 - 5)^5 + 3t \cdot 20t(2t^2 - 5)^4
\]
5. **Simplify the Expression:**
\[
\frac{dy}{dt} = 3(2t^2 - 5)^5 + 60t^2(2t^2 - 5)^4
\]
This expression gives the rate of change of \( y \) with
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