Are you asking how to prove that the derivative of \(\displaystyle e^x\) is just \(\displaystyle e^x[/itex] itself? That depends upon exactly how you define [itex]e^x[/itex].
One way to do this is to look at the general \(\displaystyle y= a^x\) where a is a fixed positive number. \(\displaystyle f(x+h)= a^{x+y}= a^xa^h\). So \(\displaystyle \frac{f(x+h)- f(x)}{h}= \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= \frac{a^h- 1}{h}a^x\). That is just \(\displaystyle a^x\) times a function of h only. The derivative of that is the limit of that as h goes to 0: \(\displaystyle \left(\lim_{h\to 0}\frac{a^h- 1}{h}\right)a^x\), a constant times \(\displaystyle a^x\). We define "e" to be that value of a such that the limit is 1.
Another way to do that is to start by defining \(\displaystyle ln(x)= \int_1^x \frac{dt}{t}\). From that definition, we can prove all of the basic properties, in particular, that ln(x) is an increasing function from \(\displaystyle (0, \infty)\) to the set of all real numbers, and so has an inverse from the set of all real number to the interval \(\displaystyle (0, \infty)\). Clearly the derivative of ln(x), by the "Fundamental Theorem of Calculus", is 1/x which is positive for all x in \(\displaystyle (0, \infty)\). So if we define "exp(x)" as the inverse of ln(x), y= exp(x) gives x= ln(y). Then dx/dy= 1/y and then dy/dx= y= exp(x).
Yes, you would use the product rule: \(\displaystyle (7x)(e^x)'+ (7x)'(e^x)= 7xe^x+ 7e^x= 7(x+ 1)e^x\). But if you already know that the derivative of \(\displaystyle xe^x\) is \(\displaystyle (x+ 1)e^x\) you can just multiply that by 7. Normally students learn that, for any constant, C, the derivative of Cf(x) is Cf'(x) before they learn the product rule!\)
I understand the stuff regarding \(\displaystyle 7ex^{x}\), but the other stuff will take some time to absorb.\)