# What is the implicit derivative of 1= (xy+y)^2-e^(xy) ?

Aug 22, 2016

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

or,

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

#### Explanation:

We rewrite the given eqn. as $1 + {e}^{x y} = {y}^{2} {\left(x + 1\right)}^{2}$.

Diff.ing both sides w.r.t. $x$, we get,

$\left(1 + {e}^{x y}\right) ' = \left\{{y}^{2} {\left(x + 1\right)}^{2}\right\} '$.

$\therefore 1 ' + \left\{{e}^{x y}\right\} ' = \left({y}^{2}\right) \left\{{\left(x + 1\right)}^{2}\right\} ' + {\left(x + 1\right)}^{2} \left({y}^{2}\right) '$.

$\therefore 0 + \left\{{e}^{x y}\right\} ' = {y}^{2} \left\{2 \left(x + 1\right) \left(x + 1\right) '\right\} + {\left(x + 1\right)}^{2} \left(2 y y '\right)$.

$\therefore {e}^{x y} \left(x y\right) ' = 2 {y}^{2} \left(x + 1\right) \left(x ' + 1 '\right) + 2 y y ' {\left(x + 1\right)}^{2}$.

$\therefore {e}^{x y} \left\{x y ' + y x '\right\} = 2 {y}^{2} \left(x + 1\right) \left(1 + 0\right) + 2 y y ' {\left(x + 1\right)}^{2}$.

$\therefore \left(x y ' + y\right) {e}^{x y} = 2 {y}^{2} \left(x + 1\right) + 2 y y ' {\left(x + 1\right)}^{2}$

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

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

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

This can further be simplified by replacing ${e}^{x y}$ by ${y}^{2} {\left(x + 1\right)}^{2} - 1$

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

Enjoy Maths.!