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How does one prove the following result regarding the nth derivative.

For $y= \left ( x^{2} +1\right )^{n}$ prove that $y_{2n} = 2n!$, where $y_{2n} $ represents the $2n^{th}$ derivative.

The main task at hand being generalizing the expression for first, second, and third derivative, which I am unable to do.

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Note that $(x^2+1)^n=x^{2n}+$lower terms. – Hagen von Eitzen Oct 30 '13 at 13:47
@Hagen Well thank you. That indeed helps. Wondering now why I did not use Binomial Expansion. – wamiq reyaz Oct 30 '13 at 13:48

Proving the formula for $y^{(2n)}$ doesn't really involve doing anything smart about the "lower" derivatives.

You just note that

$$ y = (x^2+1)^n = x^{2n} + P(x), $$ where $P(x)$ has degree at most $2n-1$. It means that $2n$'th derivative of $P(x)$ is $0$, so $y^{(2n)} = (x^{2n})^{(2n)} = (2n)!$.

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Hagen above said the same thing. I was foolish. – wamiq reyaz Oct 30 '13 at 13:51

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