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

Mar 6, 2018

${y}^{'} = \frac{\sin y - 6 x y + \frac{1}{x - y} ^ 2}{\left(\frac{1}{x - y} ^ 2 - 4 y + 3 {x}^{2} - x \cos y\right)}$

#### Explanation:

$2 {y}^{2} - 3 {x}^{2} y + x \sin y = \frac{1}{x - y}$

We use the chain rule to implicitly differentiate:

$4 y {y}^{'} - 6 x y - 3 {x}^{2} {y}^{'} + \sin y + x {y}^{'} \cos y = - \frac{1}{x - y} ^ 2 + \frac{{y}^{'}}{x - y} ^ 2$

Combine terms with y':

$\left(4 y - 3 {x}^{2} + x \cos y\right) {y}^{'} + \sin y - 6 x y = \frac{{y}^{'} - 1}{x - y} ^ 2$

Get rid of fraction on RHS:

${\left(x - y\right)}^{2} \left(4 y - 3 {x}^{2} + x \cos y\right) {y}^{'} + {\left(x - y\right)}^{2} \sin y - 6 x y {\left(x - y\right)}^{2} = {y}^{'} - 1$

Move all y' terms to RHS:

${\left(x - y\right)}^{2} \sin y - 6 x y {\left(x - y\right)}^{2} + 1 = {y}^{'} - {\left(x - y\right)}^{2} \left(4 y - 3 {x}^{2} + x \cos y\right) {y}^{'}$

Factor out y':

${\left(x - y\right)}^{2} \sin y - 6 x y {\left(x - y\right)}^{2} + 1 = {y}^{'} \left(1 - {\left(x - y\right)}^{2} \left(4 y - 3 {x}^{2} + x \cos y\right)\right)$

Solve for y':
${y}^{'} = \frac{{\left(x - y\right)}^{2} \sin y - 6 x y {\left(x - y\right)}^{2} + 1}{\left(1 - {\left(x - y\right)}^{2} \left(4 y - 3 {x}^{2} + x \cos y\right)\right)}$

Simplify:
${y}^{'} = \frac{\sin y - 6 x y + \frac{1}{x - y} ^ 2}{\left(\frac{1}{x - y} ^ 2 - \left(4 y - 3 {x}^{2} + x \cos y\right)\right)}$

Solution:
${y}^{'} = \frac{\sin y - 6 x y + \frac{1}{x - y} ^ 2}{\left(\frac{1}{x - y} ^ 2 - 4 y + 3 {x}^{2} - x \cos y\right)}$