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d²w dz² w=11z²e² Find for the given function.

d²w dz² w=11z²e² Find for the given function.

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 \(\dfrac{d^2 w}{dz^2}\) for the given function. \[ w = 11z^2 e^z \] ### Explanation Given a function \( w = 11z^2 e^z \), the task is to find the second derivative of \( w \) with respect to \( z \). ### Steps to Solve 1. **First Derivative Calculation:** - Use the product rule for differentiation: \((uv)' = u'v + uv'\). For \( w = 11z^2 e^z \): - Let \( u = 11z^2 \) and \( v = e^z \). - Calculate \( u' \) and \( v' \): \[ u' = \dfrac{d}{dz} (11z^2) = 22z \] \[ v' = \dfrac{d}{dz} (e^z) = e^z \] - Use the product rule: \[ \dfrac{dw}{dz} = u'v + uv' \] \[ \dfrac{dw}{dz} = (22z)e^z + (11z^2)e^z \] \[ \dfrac{dw}{dz} = 22ze^z + 11z^2e^z \] 2. **Simplify the First Derivative:** \[ \dfrac{dw}{dz} = 11(2z + z^2)e^z \] 3. **Second Derivative Calculation:** - Use the product rule again for the first derivative expression \(\dfrac{dw}{dz} = 11(2z + z^2)e^z \). - Let \( u = 11(2z + z^2) \) and \( v = e^z \). - Calculate \( u' \) and \( v' \): \[ u' = \dfrac{d}{dz} [11(2z + z^2)] = 11(2 + 2z) \] \[ v' = e^z \] - Use the product rule: \[ \dfrac{d^2w}{dz^2} = u'v + uv' \]
Transcribed Image Text:### Problem Statement Find \(\dfrac{d^2 w}{dz^2}\) for the given function. \[ w = 11z^2 e^z \] ### Explanation Given a function \( w = 11z^2 e^z \), the task is to find the second derivative of \( w \) with respect to \( z \). ### Steps to Solve 1. **First Derivative Calculation:** - Use the product rule for differentiation: \((uv)' = u'v + uv'\). For \( w = 11z^2 e^z \): - Let \( u = 11z^2 \) and \( v = e^z \). - Calculate \( u' \) and \( v' \): \[ u' = \dfrac{d}{dz} (11z^2) = 22z \] \[ v' = \dfrac{d}{dz} (e^z) = e^z \] - Use the product rule: \[ \dfrac{dw}{dz} = u'v + uv' \] \[ \dfrac{dw}{dz} = (22z)e^z + (11z^2)e^z \] \[ \dfrac{dw}{dz} = 22ze^z + 11z^2e^z \] 2. **Simplify the First Derivative:** \[ \dfrac{dw}{dz} = 11(2z + z^2)e^z \] 3. **Second Derivative Calculation:** - Use the product rule again for the first derivative expression \(\dfrac{dw}{dz} = 11(2z + z^2)e^z \). - Let \( u = 11(2z + z^2) \) and \( v = e^z \). - Calculate \( u' \) and \( v' \): \[ u' = \dfrac{d}{dz} [11(2z + z^2)] = 11(2 + 2z) \] \[ v' = e^z \] - Use the product rule: \[ \dfrac{d^2w}{dz^2} = u'v + uv' \]
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