How do you implicitly differentiate # 2y^2 - 3x^2y + x siny= 1/(x-y)#?

1 Answer
Mar 6, 2018

#y^(')=(siny-6xy+(1)/(x-y)^2)/(((1)/(x-y)^2-4y+3x^2-xcosy))#

Explanation:

We start with:

#2y^2 - 3x^2y + x siny= 1/(x-y)#

We use the chain rule to implicitly differentiate:

#4yy^(') - 6xy - 3x^2y^(')+ siny + xy^(')cosy = -(1)/(x-y)^2 + (y^('))/(x-y)^2#

Combine terms with y':

#(4y - 3x^2+ xcosy)y^(') +siny-6xy= (y^(') - 1)/(x-y)^2#

Get rid of fraction on RHS:

#(x-y)^2(4y-3x^2+xcosy)y^(') + (x-y)^2siny-6xy(x-y)^2 = y^(') - 1#

Move all y' terms to RHS:

#(x-y)^2siny-6xy(x-y)^2+1=y^(') - (x-y)^2(4y-3x^2+xcosy)y^(')#

Factor out y':

#(x-y)^2siny-6xy(x-y)^2+1=y^(')(1-(x-y)^2(4y-3x^2+xcosy))#

Solve for y':
#y^(')=((x-y)^2siny-6xy(x-y)^2+1)/((1-(x-y)^2(4y-3x^2+xcosy)))#

Simplify:
#y^(')=(siny-6xy+(1)/(x-y)^2)/(((1)/(x-y)^2-(4y-3x^2+xcosy)))#

Solution:
#y^(')=(siny-6xy+(1)/(x-y)^2)/(((1)/(x-y)^2-4y+3x^2-xcosy))#