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

Apr 14, 2017

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

#### Explanation:

Take the derivative of the equation except for every time you differentiate $y$, it becomes $\frac{\mathrm{dy}}{\mathrm{dx}}$. You will also have to use the product rule.

Note: ${e}^{x - 2 y} = \left({e}^{x}\right) \left({e}^{-} \left(2 y\right)\right)$. So the equation is actually:

$y = x {y}^{2} + {x}^{2} \left({e}^{x}\right) \left({e}^{-} \left(2 y\right)\right)$ so you will have to use the product rule twice for the second term:

dy/dx=x(2ydy/dx)+y^2(1)+(x^2(e^xe^-(2y))' + (e^x)(e^-(2y))(2x)

Use the product rule again:

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

Distribute the x^2:

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

Separate the terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Factor out the $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

Divide both sides by $\left(1 - 2 x y + 2 {x}^{2} {e}^{x - 2 y}\right)$:

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

Apr 14, 2017

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

$\implies y ' = {y}^{2} + 2 x y y ' + 2 x {e}^{x - 2 y} + {x}^{2} \left(\textcolor{red}{1} - 2 y '\right) {e}^{x - 2 y}$

$\implies y ' \left(1 - 2 x y + 2 {x}^{2} {e}^{x - 2 y}\right) = {y}^{2} + \left(\textcolor{red}{{x}^{2} +} 2 x\right) {e}^{x - 2 y}$

$\implies y ' = \frac{{y}^{2} + \left(\textcolor{red}{{x}^{2} +} 2 x\right) {e}^{x - 2 y}}{1 - 2 x y + 2 {x}^{2} {e}^{x - 2 y}}$

Apr 14, 2017

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

#### Explanation:

Using the Implicit Function Theorem we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\mathrm{dx}}}{\frac{\partial f}{\mathrm{dy}}}$

Where:

$f \left(x , y\right) = 0$

We have:

$\setminus \setminus \setminus \setminus \setminus y = x {y}^{2} + {x}^{2} {e}^{x - 2 y}$
$\therefore y = x {y}^{2} + {x}^{2} {e}^{x} {e}^{- 2 y}$

Let:

$f \left(x , y\right) = x {y}^{2} + {x}^{2} {e}^{x} {e}^{- 2 y} - y$

Then the partial derivatives are:

$\frac{\partial f}{\partial x} = {y}^{2} + \left({x}^{2}\right) \left({e}^{x} {e}^{- 2 y}\right) + \left(2 x\right) \left({e}^{x} {e}^{- 2 y}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {y}^{2} + \left({x}^{2}\right) \left({e}^{x - 2 y}\right) + \left(2 x\right) \left({e}^{x - 2 y}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {y}^{2} + \left({x}^{2} + 2 x\right) {e}^{x - 2 y}$

And:

$\frac{\partial f}{\partial y} = 2 x y - 2 {x}^{2} {e}^{x} {e}^{- 2 y} - 1$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 x y - 2 {x}^{2} {e}^{x - 2 y} - 1$

And so:

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

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