A function f : R ® R satisfies the equation

Given,

f : R ® R satisfies the equation f ( x + y) = f (x) f (y) for all x, y ÎR, f (x) ¹ 0

Let us take any point x = 0 at which the function f(x) is differentiable.

So, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$2=\lim _{h \rightarrow 0} \frac{f(0) \cdot f(h)-f(0)}{h} \quad[\because f(0)=f(h)]$ .....$\ldots(i)$

$\Rightarrow 2=\lim _{h \rightarrow 0} \frac{f(0)[f(h)-1]}{h}$

Now, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h} \quad[\because f(x+y)=f(x) \cdot f(y)]$

$=\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=2 f(x)$ from eqn. (i)

Therefore, f’(x) = 2f(x).