Derivative of lne^x = 1, however, confused by this person's method of solving:

Those steps are correct. I think you just need to review the derivatives of ln and e.

Let f(x) be a function of x. The derivative of ln(f(x)) is f'(x)/f(x).
For example, consider ln(x^2). Here x^2 is f(x), so f'(x) = 2x.
d/dx ln(x^2) = (2x)/(x^2) = (2x)/(xx) ... You can cancel an x... = 2/x.

So for ln(e^x), f(x) = e^x. Thus,
D/dx ln(e^x) = (d/dx e^x)/(e^x). At this point you need to know the derivative of e^x, which is just e^x.
So then you get (e^x)/(e^x) = 1 since the numerator and denominator are equal.

the derivative of ln x = 1/x. and derivative of ln (P(x))= P'(x)/P(x). both of those you just kinda know. you can verify with definition of the derivative.

(ln(y))' = y'/y.
So (ln(e^x))' = (e^x)/(e^x) = 1.

A simpler way to show this is:
ln(e^x) = x => (ln(e^x))' = (x)' = 1.

lne^x = x

so derivative = 1

Why do all that? ln(x) and e^x are inverse functions, so ln(e^x) is the same function as x, and dx/dx = 1.

When you have a ln and an e together, they cancel each other out so you're left with the x.