Question

How do you find the instantaneous rate of change of `lnx`? please explain the steps

Answer 1

The underlying idea is this, courtesy of Jacob Bernoulli :

• `lim_(sigma to oo ) (1+1/sigma )^sigma = e`

So, from algebra:

`y = ln x`

`y +delta y = ln (x + delta x)`

`delta y = (y + delta y ) - y = ln (x + delta x) - ln x`

`= ln ((x + delta x)/x) = ln (1 + (delta x)/x)`

`(delta y)/(delta x) = 1/(delta x) ln (1 + (delta x)/x)`

`= 1/(delta x) (delta x)/(x) x/(delta x) ln (1 + (delta x)/x)`

`= 1/x ln (1 + (delta x)/x)^(x/(delta x))`

Let `sigma = x/(delta x)` and you are done