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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![### Calculating Higher Order Derivatives
**Problem Statement:**
If \( y = e^{2x} + e^{3x} \),
then find the value of \(\frac{d^3 y}{dx^3}\) at \( x = 0 \).
**Solution:**
To solve this, we need to calculate the third derivative of \( y \) with respect to \( x \) and evaluate it at \( x = 0 \).
1. **First Derivative (\( \frac{dy}{dx} \)):**
Given \( y = e^{2x} + e^{3x} \),
\[
\frac{dy}{dx} = 2e^{2x} + 3e^{3x}
\]
2. **Second Derivative (\( \frac{d^2y}{dx^2} \)):**
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}(2e^{2x} + 3e^{3x}) = 4e^{2x} + 9e^{3x}
\]
3. **Third Derivative (\( \frac{d^3y}{dx^3} \)):**
\[
\frac{d^3y}{dx^3} = \frac{d}{dx}(4e^{2x} + 9e^{3x}) = 8e^{2x} + 27e^{3x}
\]
4. **Evaluate at \( x = 0 \):**
\[
\frac{d^3y}{dx^3} \bigg|_{x=0} = 8e^{0} + 27e^{0} = 8 + 27 = 35
\]
Therefore, the answer is \(\boxed{35}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd94ee947-1f00-4d5c-8d00-18458a43a16b%2Fcc6bed3f-2430-43b3-968f-805634732f91%2Fbbhl978_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating Higher Order Derivatives
**Problem Statement:**
If \( y = e^{2x} + e^{3x} \),
then find the value of \(\frac{d^3 y}{dx^3}\) at \( x = 0 \).
**Solution:**
To solve this, we need to calculate the third derivative of \( y \) with respect to \( x \) and evaluate it at \( x = 0 \).
1. **First Derivative (\( \frac{dy}{dx} \)):**
Given \( y = e^{2x} + e^{3x} \),
\[
\frac{dy}{dx} = 2e^{2x} + 3e^{3x}
\]
2. **Second Derivative (\( \frac{d^2y}{dx^2} \)):**
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}(2e^{2x} + 3e^{3x}) = 4e^{2x} + 9e^{3x}
\]
3. **Third Derivative (\( \frac{d^3y}{dx^3} \)):**
\[
\frac{d^3y}{dx^3} = \frac{d}{dx}(4e^{2x} + 9e^{3x}) = 8e^{2x} + 27e^{3x}
\]
4. **Evaluate at \( x = 0 \):**
\[
\frac{d^3y}{dx^3} \bigg|_{x=0} = 8e^{0} + 27e^{0} = 8 + 27 = 35
\]
Therefore, the answer is \(\boxed{35}\).
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