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OCW Scholar - Single Variable Calculus 16 Online
OpenStudy (anonymous):

how D^2 = n(n-1)x^(n-2)?

OpenStudy (anonymous):

The expression you have to the right of the equals sign is the second derivative of x^n, which could be written as follows\[D^{(2)}x ^{n}=n(n-1)x ^{n-2}\]In this expression, D with superscript 2 is not D squared, but instead indicates the second derivative. When you take the first derivative you get\[(x ^{n})'=nx ^{n-1}\]and when you take the second derivative you get\[(x ^{n})''=(nx ^{n-1})'=n(n-1)x ^{n-2}\]

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