Question

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

Answer 1

`y'=(x^2-xy+y^2)/(xy)`

Explanation:

`Ln(x-y)=x/y`

`(1-y')/(x-y)=(1*y-xy')/x^2`

`(1-y')/(x-y)=(y-xy')/x^2`

#(1-y')x^2=(x-y)(y-xy')

`x^2-x^2*y'=xy-x^2*y'-y^2+xyy'`

`x^2-xy+y^2=xy*y'`

`y'=(x^2-xy+y^2)/(xy)`

Answer 2

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

Explanation:

`"differentiate "ln(x-y)" using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`rArrd/dx(ln(x-y))`

`=1/(x-y)xxd/dx(x-y)=(1-dy/dx)/(x-y)`

`"differentiate "x/y" using the "color(blue)"quotient rule"`

`"given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

`rArrd/dx(x/y)=(y-xdy/dx)/y^2`

`"equating the derivates on both sides"`

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

`rArry^2-y^2dy/dx=xy-x^2dy/dx-y^2+xydy/dx`

`rArrx^2dy/dx-xydy/dx-y^2dy/dx=xy-y^2-y^2`

`rArrdy/dx(x^2-xy-y^2)=xy-2y^2`

`rArrdy/dx=(xy-2y^2)/(x^2-xy-y^2)`