# Find dy/dx for the problem below bu using IMPLICIT DIFFERENTIATION only. x^y=y^x

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### 1 Answer

You should take logarithms both sides such that:

`ln(x^y) = ln(y^x)`

Using logarithmic properties yields:

`y*ln x = x*ln y => (ln y)/y = (ln x)/x`

You should differentiate with respect to x both sides such that:

`((ln y)'*y - (ln y)*y')/(y^2) = ((ln x)'*x - (ln x)*x')/(x^2)`

`((1/y)*y(dy)/(dx) - (ln y)*(dy)/(dx))/(y^2) = (1/x*x - (ln x))/(x^2)`

Factoring out `(dy)/(dx)` to the left side yields:

`(dy)/(dx)(1 - ln y)/(y^2) = (1 - ln x)/(x^2)`

`(dy)/(dx) = ((1 - ln x)/(x^2))/((1 - ln y)/(y^2))`

`(dy)/(dx) = (y^2(1 - ln x))/(x^2(1 - ln y))`

**Hence, differentiating the given function yields `(dy)/(dx) = (y^2(1 - ln x))/(x^2(1 - ln y)).` **