Find the instantaneous rate of change of h(x): Write the full coluti - (x3 2 5+1)8 at x = 0.
Find the instantaneous rate of change of h(x): Write the full coluti - (x3 2 5+1)8 at x = 0.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find the instantaneous rate of change of \( h(x) = \frac{2}{(x^3 - 5x + 1)^8} \) at \( x = 0 \).
**Solution:**
To find the instantaneous rate of change of the function \( h(x) \) at \( x = 0 \), we need to compute the derivative \( h'(x) \) and then evaluate it at \( x = 0 \).
The derivative can be found using the chain rule. Given the function \( h(x) = [f(x)]^g \), where \( f(x) = \frac{2}{x^3 - 5x + 1} \) and \( g = -8 \), we apply the chain rule:
\[ h'(x) = \frac{d}{dx}\left\{ \left( \frac{2}{x^3 - 5x + 1} \right)^{-8} \right\} \]
For the term \( \left( \frac{2}{x^3 - 5x + 1} \right)^{-8} \):
1. Let \( u = x^3 - 5x + 1 \).
2. Then, \( h(x) = 2u^{-8} \).
Taking the derivative with respect to \( u \):
\[ \frac{d}{dx} (2u^{-8}) = 2(-8)u^{-9} \cdot \frac{du}{dx} \]
Where \( \frac{du}{dx} \) is the derivative of \( u \):
\[ \frac{d}{dx} (x^3 - 5x + 1) = 3x^2 - 5 \]
Thus:
\[ h'(x) = -16u^{-9} \cdot (3x^2 - 5) \]
\[ = -16 (x^3 - 5x + 1)^{-9} \cdot (3x^2 - 5) \]
Evaluating this at \( x = 0 \):
\[ h' (0) = -16(0^3 - 5(0) + 1)^{-9} \cdot (3(0)^2 - 5) \]
\[ = -16(1)^{-9} \cdot (-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a780900-9b69-4fe9-9f43-185066935266%2F2119ec8c-3d71-4a9d-9763-f6659f04bd02%2Fxbhj7uh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the instantaneous rate of change of \( h(x) = \frac{2}{(x^3 - 5x + 1)^8} \) at \( x = 0 \).
**Solution:**
To find the instantaneous rate of change of the function \( h(x) \) at \( x = 0 \), we need to compute the derivative \( h'(x) \) and then evaluate it at \( x = 0 \).
The derivative can be found using the chain rule. Given the function \( h(x) = [f(x)]^g \), where \( f(x) = \frac{2}{x^3 - 5x + 1} \) and \( g = -8 \), we apply the chain rule:
\[ h'(x) = \frac{d}{dx}\left\{ \left( \frac{2}{x^3 - 5x + 1} \right)^{-8} \right\} \]
For the term \( \left( \frac{2}{x^3 - 5x + 1} \right)^{-8} \):
1. Let \( u = x^3 - 5x + 1 \).
2. Then, \( h(x) = 2u^{-8} \).
Taking the derivative with respect to \( u \):
\[ \frac{d}{dx} (2u^{-8}) = 2(-8)u^{-9} \cdot \frac{du}{dx} \]
Where \( \frac{du}{dx} \) is the derivative of \( u \):
\[ \frac{d}{dx} (x^3 - 5x + 1) = 3x^2 - 5 \]
Thus:
\[ h'(x) = -16u^{-9} \cdot (3x^2 - 5) \]
\[ = -16 (x^3 - 5x + 1)^{-9} \cdot (3x^2 - 5) \]
Evaluating this at \( x = 0 \):
\[ h' (0) = -16(0^3 - 5(0) + 1)^{-9} \cdot (3(0)^2 - 5) \]
\[ = -16(1)^{-9} \cdot (-
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