A function $f:R \to R$ satisfies the equation $f(x + y) = f(x) \cdot f(y)$ for all $x,y \in R,$ $f(x) \ne 0$. Suppose that the function is differentiable at $x = 0$ and ${f^\prime }(0) = 2$, then prove that ${f^\prime }(x) = 2f(x)$

Let $f:R \to R$ satisfies the equation $f(x + y) = f(x) \cdot f(y),\,\,\forall x,y \in R,f(x) \ne 0$

Let $f(x)$ is differentiable at $x = 0$ and ${f^\prime }(0) = 2$.....(i).

$\Rightarrow$ ${f^\prime }(0) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f(0)}}{{x - 0}}$

$\Rightarrow$ $2 = \mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f(0)}}{x}$

$\Rightarrow$ $2 = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{{0 + h}}$

$\Rightarrow$ $2 = \mathop {\lim }\limits_{h \to 0} \frac{{f(0) \cdot f(h) - f(0)}}{h}$

$\Rightarrow$ $2 = \mathop {\lim }\limits_{h \to 0} \frac{{f(0)[f(h) - 1]}}{h}$

Also, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{f(x) \cdot f(h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{f(x)[f(h) - 1]}}{h} = 2f(x)$ [using Eq. (i)]

$\Rightarrow {f^\prime }(x) = 2f(x)$

Hence proved.

Differentiate each of the following w.r.t. x(Exercise 25 to 43)