# How do you implicitly differentiate -1=xy^2+2x^2y-e^ysin(3x+y) ?

##### 1 Answer

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

#### Explanation:

From the given $- 1 = x {y}^{2} + 2 {x}^{2} y - {e}^{y} \cdot \sin \left(3 x + y\right)$
Differentiate both sides of the equation with respect to $x$

it goes like this

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

$0 = x \cdot 2 y y ' + {y}^{2} \cdot 1 + 2 \left[{x}^{2} y ' + y \cdot 2 x\right] - \left[{e}^{y} \cdot y ' \cdot \sin \left(3 x + y\right) + {e}^{y} \cdot \cos \left(3 x + y\right) \cdot \left(3 + y '\right)\right]$

simplification, it follows

$0 = 2 x y y ' + {y}^{2} + 2 {x}^{2} y ' + 4 x y - {e}^{y} \sin \left(3 x + y\right) y ' - 3 {e}^{y} \cos \left(3 x + y\right) - {e}^{y} \cos \left(3 x + y\right) y '$

collecting terms with $y '$ on one side and transposing the rest of the terms on the other side;

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

Solve for $y '$ by dividing both sides now by $\left[2 x y + 2 {x}^{2} - {e}^{y} \sin \left(3 x + y\right) - {e}^{y} \cos \left(3 x + y\right)\right]$

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