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

Feb 25, 2017

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

OR

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

#### Explanation:

To avoid $- v e$signs, let us rewrite the eqn. as,

${y}^{2} x + y {e}^{x y} = x y + 1.$

Diff.ing term-wise, using the usual Rules of Diffn., we get,

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

$= \frac{d}{\mathrm{dx}} \left(x y\right) + 0.$

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

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

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

;. {2xy+xye^(xy)+e^(xy)-x}dy/dx=y-y^2-y^2e^(xy).

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

This Answer is fairly reasonable, but, if we replace ${e}^{x y}$ from this, we can have better Answer as shown below :

$N R . = y \left\{1 - y - \left(x y + 1 - {y}^{2} x\right)\right\} = y \left(1 - y - x y - 1 + {y}^{2} x\right) = y \left(x {y}^{2} - x y - y\right) = {y}^{2} \left(x y - x - 1\right) .$

$D R . = 2 x y - x + x \left(x y + 1 - {y}^{2} x\right) + \frac{x y + 1 - {y}^{2} x}{y} ,$

$= \frac{2 x {y}^{2} - x y + {x}^{2} {y}^{2} + x y - {y}^{3} {x}^{2} + x y + 1 - {y}^{2} x}{y}$

$= \frac{x y + x {y}^{2} + {x}^{2} {y}^{2} - {x}^{2} {y}^{3} + 1}{y} .$

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