How do you implicitly differentiate #xy^2- yln(x-y)^2= 4-x#?

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Answer 1 · 2016-01-07T01:17:50

Remember to use the chain rule and product rule when differentiating the y variable ...

#xy^2- yln(x-y)^2= 4-x#

Now, differentiate ...

#y^2+x(2y)y'-ln(x-y)^2-y/(x-y)^2(2)(x-y)(1-y')=-1#

Next simplify and solve for #y'# ...

#y'[(2xy)+(2y)/(x-y)]=ln(x-y)^2+(2y)/(x-y)-1#

#y'=[ln(x-y)^2+(2y)/(x-y)-1]/[(2xy)+(2y)/(x-y)]#

hope that helped