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Prove that if n = nxn-1 holds integer value of -n, which means that for a positive integer value n, then it also holds for a negative Hint: quotient rule! -n n =-nx-n-1.

Prove that if n = nxn-1 holds integer value of -n, which means that for a positive integer value n, then it also holds for a negative Hint: quotient rule! -n n =-nx-n-1.

Advanced Engineering Mathematics
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 4:** Prove that if \(\frac{d}{dx} x^n = nx^{n-1}\) holds for a positive integer value of \(n\), then it also holds for a negative integer value of \(n\). This means that \(\frac{d}{dx} x^n = nx^{n-1}\). *Hint: Quotient rule!*
Transcribed Image Text:**Problem 4:** Prove that if \(\frac{d}{dx} x^n = nx^{n-1}\) holds for a positive integer value of \(n\), then it also holds for a negative integer value of \(n\). This means that \(\frac{d}{dx} x^n = nx^{n-1}\). *Hint: Quotient rule!*
Expert Solution
Step 1

We have given that ddxxn=nxn-1 for any positive integer n.

We need to show that, ddxx-n=-nx-n-1.

We know the quotient rule of derivative,

ddxuxvx=u'xvx-vxu'xvx2

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