Connecting ln(x) and 1/x

I have more from Strang. We're going to look at the the limit of Δy/Δx in the context of the exponential. Following the book, we use the symbol h for Δx, the small change in x. So we have:

dy/dx = lim(h->0) [y(x+h) - y(x)] / h]

In this case, y(x) = b^x so we want

dy/dx = lim(h->0) [ bx+h - bx ] / h ]

As he says, the key idea is that we can split the first part

bx+h = bx bh

and then factor out b^x and move it outside the limit:

dy/dx = bx lim(h->0) [bh - 1] / h ]

The term in the limit is something. It depends on what the limit converges too. However, the important thing is that it is not a variable but a constant. Thus,

dy/dx = c bx

This is the equation that I used in the last post. Starting from here there is a quick demonstration of something that we proved in a more roundabout fashion before. Start with the equation above and invert it:

dx/dy = 1 / c bx

but

y = bx x = logb y dx/dy = 1/cy d/dy logb(y) = 1/cy

Beautiful! Strang says "that proof was...powerful too quick." So he does the following.

y = bx f(y) = x f(bx) = x f'(bx) (c bx) = 1 # this is the chain rule f'(bx) = 1 / (c bx) # identify bx as y f'(y) = 1/cy

f(y) converts y to x, it takes the logarithm to base b of x.