Question

How do you find the derivative of `ln x / ln y = ln(x-y)`?

Answer 1

`(dy)/(dx)=(x(lny)^2-(x-y)lny)/(x(lny)^2-(x-y)lnx)xxy/x`

Explanation:

As `(lnx)/(lny)=ln(x-y)`, using chain rule and differentiating both sides

`(lnyxx1/x-lnx xx1/yxx(dy)/(dx))/(lny)^2=1/(x-y)xx(1-(dy)/(dx))`

(note that we have used quotient rule on LHS)

`(lny/x-lnx/yxx(dy)/(dx))=(lny)^2/(x-y)-(dy)/(dx)((lny)^2/(x-y))`

or `(dy)/(dx)((lny)^2/(x-y))-lnx/yxx(dy)/(dx)=(lny)^2/(x-y)-lny/x`

or `(dy)/(dx)((lny)^2/(x-y)-lnx/y)=(lny)^2/(x-y)-lny/x`

or `(dy)/(dx)=((lny)^2/(x-y)-lny/x)/((lny)^2/(x-y)-lnx/y)`

or `(dy)/(dx)=(x(lny)^2-(x-y)lny)/(x(lny)^2-(x-y)lnx)xxy/x`