How do you implicitly differentiate $2 y^{2} - 3 x^{2} y + x sin y = \frac{1}{x - y}$ $2y^2 - 3x^2y + x siny= 1/(x-y)$ ?

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Answer 1 · 2018-03-06T00:49:13

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

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

How do you implicitly differentiate 2y^2 - 3x^2y + x siny= 1/(x-y)2y2−3x2y+xsiny=1x−y 2y^2 - 3x^2y + x siny= 1/(x-y)?