How do you implicitly differentiate #ln(x/y)=x-y #?

It is

#d(lnx-lny)/dx=d(x-y)/dx=> 1/x-[(y')/y]=1-dy/dx=> dy/dx-dy/dx*[1/y]=1-1/x=> dy/dx=[1-1/x]/[1-1/y]#

To implicitly differentiate the equation ( \ln\left(\frac{x}{y}\right) = x - y ), you follow these steps:

1. Take the natural logarithm of both sides of the equation.
2. Differentiate both sides of the equation with respect to ( x ).
3. Apply the chain rule where necessary.
4. Solve for ( \frac{dy}{dx} ).

Applying these steps, you get:

[ \frac{d}{dx}\left(\ln\left(\frac{x}{y}\right)\right) = \frac{d}{dx}(x - y) ]

[ \frac{1}{\frac{x}{y}}\left(\frac{d}{dx}\left(\frac{x}{y}\right)\right) = 1 - \frac{dy}{dx} ]

[ \frac{y}{x}\left(\frac{1}{y}\frac{dx}{dx} - \frac{x}{y^2}\frac{dy}{dx}\right) = 1 - \frac{dy}{dx} ]

[ \frac{1}{x} - \frac{x}{y^2}\frac{dy}{dx} = 1 - \frac{dy}{dx} ]

[ -\frac{x}{y^2}\frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x} ]

[ \left(1 - \frac{x}{y^2}\right)\frac{dy}{dx} = 1 - \frac{1}{x} ]

[ \frac{dy}{dx} = \frac{1 - \frac{1}{x}}{1 - \frac{x}{y^2}} ]