A real-valued function f satisfies for all reals x and y the equality
f(xy)=f(x)y+xf(y)
Prove that this function satisfies for all reals x and y≠0 the equality
f(x/y)=(f(x)y-xf(y))/y2
a) Let x = y = 1.
Then f(1) = 2*f(1) from the initial equality
so f(1) = 0
b) let x = 1/y.
Then f(1) = f(1/y)y + f(y)/y from the initial equality
But f(1) = 0 from (a)
so f(1/y) = -f(y)/y^2
c) f(x/y) = f(x)/y + xf(1/y) from the initial equality
= f(x)/y - xf(y)/y^2 from (b)
= (f(x)y - xf(y))/y^2
q.e.d.
Solution