If y=(tan-1x)^2 ..(tan inverse x whole squared) , p.t. ((x^2+1)^2 )y2 + 2x(x^2+1)y1=2

if y=(tan-1x)2…(tan inverse x whole squared) , p.t. ((x2+1)2 )y2 + 2x(x2+1)y1=2

Answer : Given : y=(tan-1x)2
To Prove : ((x2 +1)2 )y2 + 2x(x2 +1)y1=2
now,
y=(tan-1x)2
diff both sides wrt to x
=> y1 = (2 tan-1x ) / ( 1 + x2 ) { d( tan -1 x) / dx = 1/ (1 + x2 ) }
=>( 1 + x2) y1 = 2 tan-1x
diff again both sides wrt x
=> ( 1 + x2 ) y2 + 2x y1= 2 /( 1 + x2 )
=> ( 1 + x2 )2 y2 + 2x ( 1+x2 )y1= 2
hence proved
here y1 = dy / dx ( single differentiation) and
y2 = d2y / dx2 ( double differentiation)