Question

How to find dy/dx of ln(xy) = e^xy

Answer 1

Chain rule and product rule.

Answer 2

\[\dfrac{d}{dx}\ln(xy) = \dfrac{d}{dx}e^{xy}\]\[\dfrac{\left(\dfrac{d}{dx}xy \right)}{xy} = \left(\dfrac{d}{dx}xy \right)e^{xy}\]\[\dfrac{1y + xy'}{xy} = (y+xy')e^{xy}\]
Now isolate y'.

Answer 3

\[\large y + xy' = xy^2e^{xy} + x^2yy'e^{xy}\]Take all term with y' to one side and all other without y' to other side.
\[\large xy' - x^2 y y' e^{xy} = xy^2e^{xy} - y\]Factor y' out.
\[\large y'(x - x^2 y e^{xy}) = xy^2e^{xy} - y\]
\[\huge\boxed{y' = \dfrac{xy^2e^{xy} - y}{x - x^2 y e^{xy}}}\]