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

Question

Paisley Johnson · Accepted Answer

The underlying idea is this, courtesy of Jacob Bernoulli :

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

Ezra Adams · Answer

undefined To find the instantaneous rate of change of ln(x), you need to find the derivative of ln(x) with respect to x, which represents the slope of the tangent line to the curve of ln(x) at any given point.

The derivative of ln(x) with respect to x is found using the chain rule. The chain rule states that if you have a composition of functions, such as ln(f(x)), the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

So, for ln(x), the derivative is:

d/dx ln(x) = 1/x

This means that the instantaneous rate of change of ln(x) at any given point x is 1 divided by x.

Therefore, to find the instantaneous rate of change of ln(x) at a specific value of x, simply plug that value into the derivative function 1/x.