Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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 $y^{(2n)} = (x^{2n})^{(2n)} = (2n)!$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.