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

##### 2 Answers

#### Explanation:

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

#### 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)#