Differentiating Exponential Functional Equation

Question

The Functional Equation satisfied by the exponential $f(x)=e^{kx}$ is of the form:

$$ f(x+y)=f(x)f(y), \quad f(0)=1, f'(0)=k $$

Show that $f'(x) = kf(x)$.

Attempt

I tried applying Chain Rule to the functional equation but it doesnt work. $$ [f(x+y)]'=f'(x)f(y)+f(x)f'(y) $$

Let $y=0$, then $f'(x)=f'(x)\cdot 1+k\cdot f(x)$, which is absurd because it is saying $$ [e^{kx}]' = [e^{kx}]'+ke^{kx} $$ So I was wondering why we cannot apply chain rule directly to the Cauchy Functional Equation? It seems that I need to show directly $$ [f(x+0)]' = f'(0)f(x) $$ but I cannot see how to obtain this from the definition of the functional equation.

Note that you're differentiating with respect to $x$, so you should treat $f(y)$ as a constant (or vice versa). — MSDG, May 05 '19 at 09:10
Another one: https://math.stackexchange.com/q/1937878/42969. — Martin R, May 05 '19 at 09:12

score 4 · Accepted Answer · answered May 05 '19 at 09:10

You actually only have to apply the definition twice and use the given assumptions in order

\begin{align*}f'(x)&=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} &(\text{definition})\\ &=\lim_{h \to 0}\frac{f(x)\cdot f(h)-f(x)}{h}&(\text{functional property})\\ &=f(x)\cdot \lim_{h \to 0}\frac{ f(h)-1}{h}&(\text{linearity})\\ &=f(x)\cdot \lim_{h \to 0}\frac{ f(h)-f(0)}{h}&(\text{given assumption})\\ &=f(x)\cdot f'(0)&(\text{definition})\\ &=f(x)\cdot k&(\text{given assumption}) \end{align*}

Differentiating Exponential Functional Equation

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