IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

caprio.nups

hw to find the nth derivative of tan ^-1 x ????

caprio.nups

hey this is nt done yaar.....there is no one 2 help me here.......................................

bhargavi

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

y_n=(-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)}

y_n=(-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.

vasanth

i agree with u

cheeeeers