How do you implicitly differentiate #xy- 3yln(x-y)= 4/x#?

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Answer 1 · 2018-04-11T18:08:26

#dy/dx = (-4/x^2-y+(3y)/(x-y))/[x-3ln(x-y)+(3y)/(x-y)] #

Differentiating both sides of

#xy- 3yln(x-y)= 4/x#

with respect to #x# gives

#y+x dy/dx -3dy/dx ln(x-y)-(3y)/(x-y)(1-dy/dx) = -4/x^2#

so

#[x-3ln(x-y)+(3y)/(x-y)] dy/dx = -4/x^2-y+(3y)/(x-y)#

So

#dy/dx = (-4/x^2-y+(3y)/(x-y))/[x-3ln(x-y)+(3y)/(x-y)] #