Finally, use the Generalized Power Rule to determine[x -x], noting that x-x is the difference of two terms.
Finally, use the Generalized Power Rule to determine[x -x], noting that x-x is the difference of two terms.
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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Question
![Finally, use the Generalized Power Rule to determine \(\frac{d}{dx}[x^9 - x]\), noting that \(x^9 - x\) is the difference of two terms.
\[
\frac{d}{dx}[(x^9 - x)^9] = 9(x^9 - x)^8 \cdot \frac{d}{dx}[x^9 - x]
\]
\[
= 9(x^9 - x)^8 \cdot \left( \frac{d}{dx}[x^9] - \frac{d}{dx}[x] \right)
\]
\[
= 9(x^9 - x)^8 \left( \text{[ ]} \right)
\]
Therefore, if \(f(x) = (x^9 - x)^9\), then we have the following result.
\[
f'(x) = \frac{d}{dx}[(x^9 - x)^9] = \text{[ ]}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd48171ba-d1cc-47b1-bf9c-8041143384f0%2F2545636e-ab65-4ea5-876c-46529ad90015%2Fdiohdqh_processed.png&w=3840&q=75)
Transcribed Image Text:Finally, use the Generalized Power Rule to determine \(\frac{d}{dx}[x^9 - x]\), noting that \(x^9 - x\) is the difference of two terms.
\[
\frac{d}{dx}[(x^9 - x)^9] = 9(x^9 - x)^8 \cdot \frac{d}{dx}[x^9 - x]
\]
\[
= 9(x^9 - x)^8 \cdot \left( \frac{d}{dx}[x^9] - \frac{d}{dx}[x] \right)
\]
\[
= 9(x^9 - x)^8 \left( \text{[ ]} \right)
\]
Therefore, if \(f(x) = (x^9 - x)^9\), then we have the following result.
\[
f'(x) = \frac{d}{dx}[(x^9 - x)^9] = \text{[ ]}
\]
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