Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
The task is to find the partial derivatives \(\partial w/\partial s\) and \(\partial w/\partial t\) using the Chain Rule. **Function:** \[ w = y^3 - 2x^2y^5 \] **Values:** \[ s = -3, \quad t = 6 \] \[ x = e^s, \quad y = e^t \] **Partial Derivatives to Find:** 1. \(\partial w/\partial s\) - Calculation: \[ \partial w/\partial s = -4 \] - The given answer (-4) is marked incorrect (indicated by a red "X"). 2. \(\partial w/\partial t\) - Calculation: \[ \partial w/\partial t = 3e^{18} - 2 \] - This result is also marked incorrect (indicated by a red "X"). **Evaluation at Given Values:** - Evaluate each partial derivative at \(s = -3\) and \(t = 6\). 1. \(\partial w/\partial s\) - Calculation: \[ \partial w/\partial s = 3e^{18} - 2 \] - This result is marked incorrect (indicated by a red "X"). 2. \(\partial w/\partial t\) - Calculation: \[ \partial w/\partial t = -3 \] - This result is marked incorrect (indicated by a red "X").
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Transcribed Image Text:The task is to find the partial derivatives \(\partial w/\partial s\) and \(\partial w/\partial t\) using the Chain Rule. **Function:** \[ w = y^3 - 2x^2y^5 \] **Values:** \[ s = -3, \quad t = 6 \] \[ x = e^s, \quad y = e^t \] **Partial Derivatives to Find:** 1. \(\partial w/\partial s\) - Calculation: \[ \partial w/\partial s = -4 \] - The given answer (-4) is marked incorrect (indicated by a red "X"). 2. \(\partial w/\partial t\) - Calculation: \[ \partial w/\partial t = 3e^{18} - 2 \] - This result is also marked incorrect (indicated by a red "X"). **Evaluation at Given Values:** - Evaluate each partial derivative at \(s = -3\) and \(t = 6\). 1. \(\partial w/\partial s\) - Calculation: \[ \partial w/\partial s = 3e^{18} - 2 \] - This result is marked incorrect (indicated by a red "X"). 2. \(\partial w/\partial t\) - Calculation: \[ \partial w/\partial t = -3 \] - This result is marked incorrect (indicated by a red "X").
Expert Solution
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Step 1: Define the problem.

To find fraction numerator partial differential straight w over denominator partial differential straight s end fraction space and space fraction numerator partial differential straight w over denominator partial differential straight t end fraction by using the appropriate Chain Rule.

Function: 

straight w equals straight y cubed minus 2 straight x squared straight y straight x equals straight e to the power of straight s straight y equals straight e to the power of straight t

Values: 

straight s equals negative 3 comma space straight t equals 6

To evaluate each partial derivate at the given values.

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