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

Mar 14, 2017

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

#### Explanation:

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

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

To differentiate the R.H.S., we use the Product Rule and the

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

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

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

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

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

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

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

Similarly, $\frac{d}{\mathrm{dx}} \left\{{e}^{x y}\right\} = {e}^{x y} \frac{d}{\mathrm{dx}} \left\{x y\right\} ,$

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

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

Altogether, by $\left(\star\right) , \left(1\right) \mathmr{and} \left(2\right) ,$

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

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

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

Enjoy Maths.!