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bhargavi (67)

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y=tan-1x       differentiating     y1=1/1+x2  =1/(X+i)(x-i)=1/2i[1/(x-i)-1/(x+i)]
now   y2=(-1)/2i[1/(x-i)2-1/(x+i)2]
         y3=(-1)(-2)/2i[1/(x-i)3-1/(x+i)3] ..........similarly
        yn=(-1)(-2)......(-[n-1])/2i[1/(x-i)n-1/(x+i)n]=(-1)n-1(n-1)!/2i[1/(x-i)n-1/(x+i)n]
let   x=rcos      1=rsin     then   r=(1+x2)1/2.........(1)
yn=(-1)n-1(n-1)!/2i{r(-n)[cos-isin](-n)-r(-n)[cos+isin](-n)}, 
   =(-1)n-1(n-1)!/2i[2isin(n)]r(-n)       {using   de-moivre's  theorem}
  =(-1)n-1(n-1)!sin(n)/(1+x2)n/2       {from  (1) }
let   me   know   if  i'm  wrong.
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