Let `f(x)=ln((1-x)/(1+x))` . Find x,y for which `f(x)+f(y)=f((x+y)/(1+xy))`

For a function f(x),f(0)=0 and f'(x)=(1)/(1+x^(2)), prove that f(x)+f(y)=f((x+y)/(1-xy))

Which of the following functions have the graph symmetrical about the origin? f(x) given be f(x)+f(y)=f((x+y)/(1-xy))f(x) given by f(x)+f(y)=f(x sqrt(1-y^(2))+y sqrt(1-x^(2)))f(x) given by f(x+y)=f(x)+f(y)AA x,y in R none of these

If f(x)=log {(1+x)/(1-x)} , then f(x)+f(y) is

Let f:R^(+)rarr R be a function which satisfies f(x)f(y)=f(xy)+2((1)/(x)+(1)/(y)+1) for x,y>0. Then find f(x)

Let f((x+y)/(2))=(f(x)+f(y))/(2),x,y in R, if f'(0)=-1, then f'((x)/(2))=

Let f(x) be differentiable function such that f((x+y)/(1-xy))=f(x)+f(y)AA x and y, if lim_(x rarr0)(f(x))/(x)=(1)/(3)thenf'(1) equals

If f(x)=cos(log x), then f(x)f(y)-(1)/(2)[f((x)/(y))+f(xy)]=

If f(x)=cos(log x), then f(x)f(y)-(1)/(2)[f((x)/(y))+f(xy)]=