Does this make sense to anyone? What advice would you give me to clarify my reasoning and explanation?
One of the really "neat" features of the exponential function: $$f(x)=e^x$$ is the fact that the derivative of the function at a particular point, say x=0, is equal to the value of the function at said point, or e^0=1, in this case.
You can see what's happening here by looking at a graph of the function and finding the point (0,1). Right there, plain as day, the slope of the tangent line at (0,1) is clearly 1.
We're told that the logarithmic function: $$g(x)=ln(x)$$ is the inverse of the exponential function.
Given the connection between these two functions, how does the "cool" feature of e^x mentioned above translate to the logarithmic function?
Well, the derivative of the logarithmic function is 1/x. On the surface, this doesn't seem to bear any resemblance to the derivative of e^x. What's going on here?
The key issue is the transposition of variables, x and y! Here's the reasoning:
If we set: $$y=e^x$$ and then we solve for x: $$x=ln(y)$$ we get a valid "inverse" function. Plug in any "y" that is on the curve of y=e^x and you'll get back the appropriate "x." The problem is, the standard practice is to have "y" as the dependent variable, not "x."
So we switch "y" with "x" and get: $$y=ln(x).$$
This "switching" action can be "seen" in the graph of the function as a reflection about the diagonal line y=x.
And here's the important bit: by swapping "y" and "x," when we're looking at the graph of y=ln(x), the "x" values represent the evaluated exponentials, which were on the y-axis before. And the "y" values correspond to the exponents. So when we're looking at the graph of y=ln(x), 1/x really corresponds to 1/"exponential value."
And THERE'S the connection between the derivatives we were looking for!