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

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Answer 1 · 2016-06-17T09:56:02

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

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

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