Given $f(x) = \exp(ax)$, then the derivative, by definition is given by $\frac{df}{dx} = a\exp(ax)$. Also, it is known that $f(0) = 1$
I want to prove that $\exp((a+b)x)$ is equal to $\exp(ax)\cdot\exp(bx)$.
So, to do this, I take the derivative of $\exp((a+b)x)$ which results in $(a+b)\exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $\exp(ax)\cdot\exp(bx)$ which results in $(a+b)\exp(ax)\cdot\exp(bx)$.
Now, I evaluate $\frac{d}{dx}$$[\exp((a+b)x) - \exp(ax)\cdot\exp(bx)]$ at $x = 0$. This results in zero.
My question now is, have I completed the proof. If yes, can someone explain why?
Any function whose derivative is 0 for all $x$ is a constant function.
So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.
So how do we find the function value? How do we use this information to show the result?