# If $Y=\tan^{-1} x$ , obtain an equation showing the relationship between $Y_{n+2}$ , $Y_{n+1}$ and $Y_{n}$.

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 :)

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

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 ?
• Name $Y=Y_0$ so $$Y_{n+1}=Y_{n}'$$ Commented Apr 6, 2017 at 12:30