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

Apr 4, 2017

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

#### Explanation:

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

We have:

${e}^{x {y}^{2} - x y} - {x}^{3} {y}^{2} + y = 2$

So Let $F \left(x , y\right) = {e}^{x {y}^{2} - x y} - {x}^{3} {y}^{2} + y - 2$; Then;

$\frac{\partial F}{\partial x} = {e}^{x {y}^{2} - x y} \left({y}^{2} - y\right) - 3 {x}^{2} {y}^{2}$

$\frac{\partial F}{\partial y} = {e}^{x {y}^{2} - x y} \left(2 x y - x\right) - 2 {x}^{3} y + 1$

And so:

 dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{{e}^{x {y}^{2} - x y} \left({y}^{2} - y\right) - 3 {x}^{2} {y}^{2}}{{e}^{x {y}^{2} - x y} \left(2 x y - x\right) - 2 {x}^{3} y + 1}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 {x}^{2} {y}^{2} - {e}^{x {y}^{2} - x y} \left({y}^{2} - y\right)}{{e}^{x {y}^{2} - x y} \left(2 x y - x\right) - 2 {x}^{3} y + 1}$