`f(x)+f(y)=f((x+y)/(1-xy))` ,for all`x,yinR`.`(xy!=1)`,and `lim_(x->0) f(x)/x=2`.Find`f(sqrt3)`and `f'(-2)`.

Explanation

f(x) + f(y) = `f((x+y)/(1-xy)) " " (1)`
Putting x = y = 0 , we get f(0) = 0
Putting y = -x , we get
f(+x) + f(-x) = f(0)
or f(-x) = -f(x) `" "` (2)
Also , `underset(x to 0)("lim") (f(x))/(x) = 2 `
Now , f'(x) `=underset(hto0)("lim")(f(x+h) - f(x))/(h) " " (3)`
`underset(h to 0)("lim") (f(x+h) + f(-x))/(h) " " `[Using (2)]
`= underset(hto0)("lim")f((x+h-x)/(1-(x+h)(-x)))/(h) " " ` [Using (1)]
`=underset(h to 0)("lim")[(f((h)/(1+x(x+h))))/(h)]`
`underset(h to 0)("lim") (f((h)/(1+xh+x^(2))))/(((h)/(1+xh + x^(2)))) xx ((1)/(1 +xh + x^(2)))`
`underset(h to0)("lim")(f((h)/(1+xh + x^(2))))/(((h)/(1 +xh + x^(2)))) xx underset(h to0) ("lim")(1)/(1 +xh+x^(2))`
= ` 2 xx (1)/(1+ x^(2)) = (2)/(1+x^(2)) " " ("Using" underset( xto 0) ("lim") *(f(x))/(x) = 2)`
Integrating both sides we get
f(x) = 2 `tan^(-1) (x) + c` , where f(0) = 0 `implies c = 0`
Thus , f(x) = `2 tan^(-1)x ` .Hence ,
`f((1)/(sqrt3)) = 2 tan^(-1) ((1)/(sqrt3)) = 2 (pi)/(6) = (pi)/(3)`
and , f'(1) = `(2)/(1+ 1^(2)) = (2)/(2) = 1 `