Question

Find nth derivative of tan^(-1) (x/a).

Answer 1

@hartnn

Answer 2

Can't observe any visible pattern from my side.
http://www.wolframalpha.com/input/?i=Table%5BD%5BArcTan%5Bx%2Fa%5D%2C+%7Bx%2C+n%7D%5D%2C+%7Bn%2C+1%2C+10%7D%5D

Answer 3

@rational

Answer 4

@nincompoop

Answer 5

http://functions.wolfram.com/ElementaryFunctions/ArcTan/20/02/

Answer 6

what on earth

Answer 7

pick the one that looks easy, we can anayze..

Answer 8

none :P

Answer 9

same here lol

Answer 10

looks wolfram is interpreting x as complex number

Answer 11

hmm :| anything better?

Answer 12

let me grab pen and paper, i feel its going to be another 3-4 pages >.<

Answer 13

hehe XD

Answer 14

\[\large \begin{align}\\

y &= \arctan(x/a) \\~\\
y_1/a &= \dfrac{1}{a^2+x^2} = \dfrac{0!}{a^2+x^2} \\~\\
y_2/a &= \dfrac{-2x}{(a^2+x^2)^2} = -\dfrac{1!. \binom{2}{1}x}{a^2+x^2} \\~\\
y_3/a &= \dfrac{2(3x^2-a^2)}{(a^2+x^2)^3} = \dfrac{2!.\left( \binom{3}{1}x^2-\binom{3}{3}a^2\right)}{(a^2+x^2)^3} \\~\\
y_4/a &= -\dfrac{24(x^3-a^2x)}{(a^2+x^2)^4} = \dfrac{3!.\left( \binom{4}{1}x^3-\binom{4}{3}a^2x\right)}{(a^2+x^2)^4} \\~\\
\cdots\\~\\

y_n/a &= \dfrac{(-1)^n(n-1)!\left(\binom{n}{1}x^{n-1}a^0 -\binom{n}{3}x^{n-3}a^2 + \cdots \right)}{(a^2+x^2)^n}

\end{align}\]

Answer 15

Im not going to lie, I have just reverse calculated it from wolfram :) you're an einstein if you make sense of above stuff ;)

Answer 16

lets let it be :P