Question

Successive differentiation.

Please help.

Answer 1

\[\rm find~the~nth~derivative~of~\tan^{-1} x\]

Answer 2

Please help. I'm learning this and I don't know how to go about

Answer 3

@Michele_Laino

Answer 4

Well, can you take the first derivative? After that, it becomes a polynomial which has a more easy to see nth form.

Answer 5

Yup...
I got the first derivative

Answer 6

I'm working on your question...

Answer 7

After that I did for yn of the first derivati

Answer 8

Is this the actual question or are you trying to find the Taylor series of arctan x?

Answer 9

this is the actual question

Answer 10

the expression is gets longer everytime because of the chain rule :(

Answer 11

:(

Answer 12

@Michele_Laino

Answer 13

Please help. I'm learning this and I don't know how to go about

Answer 14

Although circular reasoning in a sense, if you're allowed to use the Taylor series for arctan x, you basically already have all the nth derivatives there.

Answer 15

the problem is not easy, nevertheless, the subsequent series of Taylor can be useful:
\[\Large \arctan x = \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}{x^{2k + 1}}}}{{2k + 1}}} \]

Answer 16

Actually this looks promising http://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots

Answer 17

They essentially just use

\[\frac{d}{dx} \arctan x = \frac{1}{1+x^2}\] and then use the geometric series:

\[\frac{1}{1-y} = \sum_{n=0}^\infty y^n\]

with \( y=-x^2\) and from there it looks like you can kinda leap frog your way into it. Although there might be some caveats, just what it looks like from a cursory glance! :D

Answer 18

:/

Answer 19

what about taylor series u ppl were telling?

Answer 20

Since you don't need the nth derivatives to get the Taylor series, it's actually not circular reasoning to use it to find the nth derivative of arctan. Finding the nth derivative of a Taylor series is much easier since the nth derivative of any polynomial is:

\[\frac{d^n}{dx^n} x^k = \frac{k}{(k-n)!} x^{k-n}\]

Answer 21

Taylor series can be differentiated term by term, and since the series above has radius of convergence \(r=1\), then we can apply this theorem:
"If we have this power series:
\[\Large f\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \]
then the function \(f(x)\) is infinitely differentiable, inside a neighborhood of \(x=0\) and radius \(r=1\) and we can write:
\[\Large {f^{\left( k \right)}}\left( x \right) = \sum\limits_{n = k}^\infty {n\left( {n - 1} \right)...\left( {n - k + 1} \right){a_n}{x^{n - k}}} \]

Answer 22

hmm
im confused

Answer 23

the series above is a powers series: where \(f(x)= \arctan x\), so using the theorem above, we can write this:
\[\large \frac{{{d^k}}}{{d{x^k}}}\left( {\arctan x} \right) = \sum\limits_{n = k}^\infty {n\left( {n - 1} \right)...\left( {n - k + 1} \right)\frac{{{{\left( { - 1} \right)}^n}}}{{2n + 1}}{x^{2n + 1 - k}}} \]

Answer 24

by using taylor series i can diff n times right?

Answer 25

yes! inifinite times

Answer 26

okay :)
so the terms will be infinte :)

Answer 27

yes! Each derivative is expressed as an infinite powers series, as we can see from the formula above

Answer 28

It is all that I know, in order to solve your question

Answer 29

okay
tysm

Answer 30

ty @Empty @Michele_Laino @baru

Answer 31

thanks!! :) @rvc