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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![### Calculus Problem: Derivatives
#### Problem Statement:
If \( y = (t^2 + 6t + 5)(6t^2 + 6) \), find \(\frac{dy}{dt}\).
#### Solution:
To solve this, apply the product rule of differentiation, which states:
\[ \frac{d}{dt} [u \cdot v] = u \cdot \frac{dv}{dt} + v \cdot \frac{du}{dt} \]
Given:
\[ u = t^2 + 6t + 5 \]
\[ v = 6t^2 + 6 \]
1. First, find \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
\[ \frac{du}{dt} = \frac{d}{dt}(t^2 + 6t + 5) = 2t + 6 \]
\[ \frac{dv}{dt} = \frac{d}{dt}(6t^2 + 6) = 12t \]
2. Apply the product rule:
\[ \frac{dy}{dt} = (t^2 + 6t + 5) \cdot 12t + (6t^2 + 6) \cdot (2t + 6) \]
This simplifies further with substitution and algebraic manipulation to get the final derivative.
\[ \frac{dy}{dt} = 12t(t^2 + 6t + 5) + (6t^2 + 6)(2t + 6) \]
Substitute and expand:
\[ = 12t^3 + 72t^2 + 60t + 12t^3 + 36t + 12t^2 + 36 \]
Combine like terms:
\[ \frac{dy}{dt} = 24t^3 + 84t^2 + 96t + 36 \]
Therefore,
\[ \boxed{\frac{dy}{dt} = 24t^3 + 84t^2 + 96t + 36} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc43de746-fc85-4a6f-800b-02b60c5c9f65%2Fbce2c69f-48e6-45ac-b129-0c6b5019d723%2F8z3s4z_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Derivatives
#### Problem Statement:
If \( y = (t^2 + 6t + 5)(6t^2 + 6) \), find \(\frac{dy}{dt}\).
#### Solution:
To solve this, apply the product rule of differentiation, which states:
\[ \frac{d}{dt} [u \cdot v] = u \cdot \frac{dv}{dt} + v \cdot \frac{du}{dt} \]
Given:
\[ u = t^2 + 6t + 5 \]
\[ v = 6t^2 + 6 \]
1. First, find \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
\[ \frac{du}{dt} = \frac{d}{dt}(t^2 + 6t + 5) = 2t + 6 \]
\[ \frac{dv}{dt} = \frac{d}{dt}(6t^2 + 6) = 12t \]
2. Apply the product rule:
\[ \frac{dy}{dt} = (t^2 + 6t + 5) \cdot 12t + (6t^2 + 6) \cdot (2t + 6) \]
This simplifies further with substitution and algebraic manipulation to get the final derivative.
\[ \frac{dy}{dt} = 12t(t^2 + 6t + 5) + (6t^2 + 6)(2t + 6) \]
Substitute and expand:
\[ = 12t^3 + 72t^2 + 60t + 12t^3 + 36t + 12t^2 + 36 \]
Combine like terms:
\[ \frac{dy}{dt} = 24t^3 + 84t^2 + 96t + 36 \]
Therefore,
\[ \boxed{\frac{dy}{dt} = 24t^3 + 84t^2 + 96t + 36} \]
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