Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 37E
Related questions
Question
![### 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:
\[ \dfrac{d^2w}{dz^2} = u'v + uv' \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70018f71-7185-4cb2-be6c-94aac78b8d2f%2Ff2da5027-0d43-4fee-addd-28a50c6d7656%2F319xi8u_processed.png&w=3840&q=75)
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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