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

Question

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

Christopher Agresta · Accepted Answer

#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))#

Christopher Agresta · Answer

undefined To implicitly differentiate the given expression, follow these steps:

1. Differentiate each term of the expression with respect to $ x $.
2. Treat $ y $ as a function of $ x $ and use the chain rule when differentiating terms involving $ y $.
3. Solve for $ \frac{{dy}}{{dx}} $ after differentiating.

Differentiating each term:

$$ \frac{{d}}{{dx}}(2y^2) - \frac{{d}}{{dx}}(3x^2y) + \frac{{d}}{{dx}}(x \sin y) = \frac{{d}}{{dx}}\left(\frac{1}{{x-y}}\right) $$

Using the chain rule:

$$ \frac{{d}}{{dx}}(2y^2) = 4yy' $$

$$ \frac{{d}}{{dx}}(3x^2y) = 6xy + 3x^2y' $$

$$ \frac{{d}}{{dx}}(x \sin y) = \sin y + x \cos y \frac{{dy}}{{dx}} $$

$$ \frac{{d}}{{dx}}\left(\frac{1}{{x-y}}\right) = -\frac{1}{{(x-y)^2}}(1 - \frac{{dy}}{{dx}}) $$

Putting it all together:

$$ 4yy' - 6xy - 3x^2y' + \sin y + x \cos y \frac{{dy}}{{dx}} = -\frac{1}{{(x-y)^2}}(1 - \frac{{dy}}{{dx}}) $$

Now, solve for $ \frac{{dy}}{{dx}} $.