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This is what I did :

Since, $Y = \tan^{-1} x$, by differentiating we can get,

$Y_1 = \dfrac{1}{x^2 + 1},\\ Y_2 =\dfrac{ -2x}{(x^2 + 1)^2},\\ Y_3 = \dfrac{2 (3x^2 -1)}{ (x^2+1)^3}$

and so on...

As per the above pattern, I know the formula for $Y_n$ will have $(x^2+1)^n$ in denominator but I'm unable to figure out the numerator. Even If I get the formula for $Y_n$ , How am I supposed to proceed? Please help. Thanks.

EDIT : I solved the question by using Leibniz theorem. Thanks to all who tried to help :)

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  • $\begingroup$ Do you mean $$Y = tan^{-1} x,\\ Y_1 = \dfrac{1}{x^2 + 1},\\ Y_2 =\dfrac{ -2x}{(x^2 + 1)^2}\\, Y_3 = \dfrac{2 (3x^2 -1)}{ (x^2+1)^3}$$ ? $\endgroup$
    – Khosrotash
    Commented Apr 6, 2017 at 12:13
  • $\begingroup$ @Khosrotash yes $\endgroup$
    – user430721
    Commented Apr 6, 2017 at 12:14
  • $\begingroup$ Does the notation $Y_n$ really mean the $n$th derivative of $Y$ here? $\endgroup$
    – pjs36
    Commented Apr 6, 2017 at 12:33
  • $\begingroup$ @pjs36 yes thats exactly what it means $\endgroup$
    – user430721
    Commented Apr 6, 2017 at 12:33
  • $\begingroup$ For more terms, see opensky.ca/~jdhildeb/arctan/arctan_diff.html. $\endgroup$
    – lhf
    Commented Apr 6, 2017 at 13:16

1 Answer 1

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IF $$Y = \tan^{-1} x,\\ Y_1 = \dfrac{1}{x^2 + 1},\\ Y_2 =\dfrac{ -2x}{(x^2 + 1)^2}\\, Y_3 = \dfrac{2 (3x^2 -1)}{ (x^2+1)^3}$$ start by take derivation of $Y$ $$ Y = \tan^{-1} x\to Y'=Y_1 = \dfrac{1}{1+x^2 }$$ then try for $Y''$ and $$ Y_1 = \dfrac{1}{x^2 + 1}\to Y_1'=\dfrac{0(x^2+1)-(2x)1}{(1+x^2)^2}=Y_2=Y''$$ can you go on ?

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  • $\begingroup$ Sir actually this is what I did. Since Y = tan^{-1}x , so I took the derivatives till Y_3. I have to find a general formula for Y_n so that I can proceed ahead. Please help me obtain Y_n $\endgroup$
    – user430721
    Commented Apr 6, 2017 at 12:18
  • $\begingroup$ Do you have an answer sheet ? or multiple choice ? (to pick the answer ) $\endgroup$
    – Khosrotash
    Commented Apr 6, 2017 at 12:21
  • $\begingroup$ No sir I don't have it. $\endgroup$
    – user430721
    Commented Apr 6, 2017 at 12:23
  • $\begingroup$ Name $Y=Y_0$ so $$Y_{n+1}=Y_{n}'$$ $\endgroup$
    – Khosrotash
    Commented Apr 6, 2017 at 12:30

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