Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
### 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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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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