A real valued function satisfies the relation `f(x+y)=f(x)+f(y)+(e^x-1)(e^y-1)`, `AAx,y in R`. If f'(0) = 2, find f(x).

Explanation

We have
`f(x+y)=f(x)+f(y)+(e^(x)-1)(e^(y)-1)AAx, in R" ...(1)"`
Differentiating w.r.t. x, keeping y as constant, we get
`f(x+y)=f'(x)+e^(x)(e^(y)-1)`
Putting x=0
`f'(y)=f'(0)+e^(y)-1`
`therefore" "f'(y)=e^(y)+1" "(as f'(0)=2)`
`therefore" "f(y)=e^(y)+y+c`
Putting x = y = 0 in (1), we get
f(0)=f(0)+f(0)+0
`therefore" "f(0)=0`
In (2), putting y=0, we get
0=1 +0+c
`rArr" "c=-1`
`therefore" "f(x)=e^(x)+x-1`