Question

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

Answer 1

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

Explanation:

Prerequisites : The Usual Rules of Differentiation.

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

`:. 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)}`.