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

##### 2 Answers
Jan 25, 2018

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

#### Explanation:

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

We could proceed using the product rule. In implicit differentiation, we must also remember to use the chain rule when differentiating a $y$ since $y$ is a function of $x$. So:

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

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

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

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

Now gather all terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left side:

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

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

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

Jan 25, 2018

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

#### Explanation:

$\text{differentiate using the "color(blue)"product/chain rules}$

$\text{given "y=f(x)g(x)" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$f \left(x\right) = x - {y}^{2} \rightarrow f ' \left(x\right) = 1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$g \left(x\right) = {e}^{x y} \Rightarrow g ' \left(x\right) = {e}^{x y} \times \frac{d}{\mathrm{dx}} \left(x y\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times x} = {e}^{x y} \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times x} = x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + y {e}^{x y}$

(x-y^2)(xe^(xy)dy/dx+ye^(xy)))+e^(xy)(1-2ydy/dx)

$= {x}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + x y {e}^{x y} - x {y}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{3} {e}^{x y} + {e}^{x y} - 2 y {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}}$

$y = \left(x - {y}^{2}\right) {e}^{x y} \leftarrow \textcolor{b l u e}{\text{original question}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} - {x}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} = x y {e}^{x y} - {y}^{3} {e}^{x y} + {e}^{x y}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x y {e}^{x y} - {y}^{3} {e}^{x y} + {e}^{x y}}{1 - {x}^{2} {e}^{x y} + x {y}^{2} {e}^{x y} + 2 y {e}^{x y}}$