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Find dt y = 31(21? – 5) $ dy dt

Find dt y = 31(21? – 5) $ dy dt

Calculus: Early Transcendentals
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 \(\frac{dy}{dt}\). 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
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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