Let f : ℝ → ℝ and g : ℝ → ℝ be functions satisfying f x + y = f x + f y…

Given $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)$
Put $x=y=0$ in given relation.

$\Rightarrow f\left(0\right)=f\left(0\right)+f\left(0\right)+f^2\left(0\right)$

$\Rightarrow f\left(0\right)=0$ or $-1$

$\because f\left(x+y\right)=f\left(x\right)+f\left(y\right)+f\left(x\right)·f\left(y\right)$

$\Rightarrow \frac{f\left(x+y\right)-f\left(x\right)}{y}=\frac{f\left(y\right)\left(1+f\left(x\right)\right)}{y}$

$\Rightarrow \underset{y\to 0}{\lim }\left(\frac{f(x+y)-f\left(x\right)}{y}\right)=\underset{y\to 0}{\lim }\left(1+f\left(x\right)\right)·\frac{f\left(y\right)}{y}$

$\because \underset{x\to 0}{\lim } g\left(x\right)=\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=1$

$\Rightarrow f^{\text{'}}\left(x\right)=1+f\left(x\right)$

$\Rightarrow f^{\text{'}}\left(0\right)=1+f\left(0\right)$

$\Rightarrow f^{\text{'}}\left(0\right)=1+0$

$\Rightarrow f^{\text{'}}\left(0\right)=1$

Again $\frac{f^{\text{'}}\left(x\right)}{1+f\left(x\right)}=1\Rightarrow \int \frac{f^{\text{'}}\left(x\right)dx}{1+f\left(x\right)}dx=\int dx$

$\Rightarrow ln\left(1+f\left(x\right)\right)=x+C$

$\Rightarrow ln \left[1+f\left(x\right)\right]=x \left(\because C=0\right)$

$\Rightarrow 1+f\left(x\right)=e^x$

$\Rightarrow f\left(x\right)=e^x-1 \Rightarrow f^{\text{'}}\left(x\right)=e^x$

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(1\right)=\mathrm{e}$

Also, $f\left(x\right)$ is differentiable for every $x\in R$.

$g\left(x\right)=\frac{f\left(x\right)}{x}=\frac{e^x-1}{x} y^{\text{'}}\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{g\left(0+h\right)-g\left(0\right)}{h}$

If $g\left(0\right)=1$ then $g^{\text{'}}\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{\frac{e^h-1}{h}-1}{h}=\underset{h\to 0}{\lim }\frac{e^h-1-h}{h^2}=\frac{1}{2}$

$g^{\text{'}}\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{g\left(0-h\right)-g\left(0\right)}{-h}=\underset{h\to 0}{\lim }\frac{e^{-h}-1}{\frac{-h}{-h}}$

$\underset{h\to 0}{\lim }\frac{e^{-h}-1+h}{h^2}=\frac{1}{2}$

$\mathrm{g}\left(\mathrm{x}\right)$ is differentiable for every $x\in R$.