Question

How do you implicitly differentiate `18=(y-x)ln(xy)`?

Answer 1

`y'=(y/x-1-ln(xy))/((x)/y-1-ln(xy))`

Explanation:

Start from the given equation and differentiate both sides of the equation with respect to x

`18=(y-x)ln (xy)`

`d/dx(18)=d/dx((y-x)*ln (yx))`

`0=(y-x)*d/dx(ln (xy))+ln (xy)*d/dx(y-x)`

`0=(y-x)*(1/(xy))*d/dx(xy)+(ln (xy))(y'-1)`

`0=(y-x)*(xy'+y*1)/(xy)+y'ln(xy)-ln(xy)`

`0=(y-x)*((y')/y+1/x)+y'ln(xy)-ln(xy)`

`0=y'+y/x-(xy')/y-1+y'ln(xy)-ln(xy)`

Transpose the terms with y'

`(xy')/y-y'-y'ln(xy)=y/x-1-ln(xy)`

factor out the y'

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

Divide both sides by `((x)/y-1-ln(xy))`

`(cancel((x)/y-1-ln(xy))*y')/cancel(((x)/y-1-ln(xy)))=(y/x-1-ln(xy))/((x)/y-1-ln(xy))`

`y'=(y/x-1-ln(xy))/((x)/y-1-ln(xy))`

God bless....I hope the explanation is useful.