How do you differentiate e^(x/y)=x-yexy=xy?

1 Answer
Jul 27, 2018

dy/dx={x-(x-y)ln(x-y)}/y, or, x/y{1-1/y*e^(x/y)}dydx=x(xy)ln(xy)y,or,xy{11yexy}.

Explanation:

Prerequisites : The Usual Rules of Differentiation.

Given that, e^(x/y)=x-yexy=xy.

:. lne^(x/y)=ln(x-y).

:. x/y=ln(x-y), or, x=yln(x-y).

:. d/dx{x}=d/dx{xln(x-y)}.

:.1=xd/dx{ln(x-y)}+ln(x-y)*d/dx{y},

i.e., 1=x*1/(x-y)*d/dx{x-y}+ln(x-y)*dy/dx,

or, 1=x/(x-y){d/dx(x)-d/dx(y)}+ln(x-y)*dy/dx.

:. 1=x/(x-y){1-dy/dx}+ln(x-y)*dy/dx.

:. 1=x/(x-y)-x/(x-y)*dy/dx+ln(x-y)dy/dx.

:. 1-x/(x-y)={ln(x-y)-x/(x-y)}dy/dx.

:. {(x-y)-x}/cancel(x-y)={(x-y)ln(x-y)-x}/cancel(x-y)*dy/dx.

:. -y={(x-y)ln(x-y)-x}dy/dx.

rArr dy/dx={x-(x-y)ln(x-y)}/y.

Since, (x-y)=e^(x/y) and ln(x-y)=x/y, we may write,

dy/dx=x/y-1/y*e^(x/y)*x/y=x/y{1-1/y*e^(x/y)}.