# How do you implicitly differentiate -1=(x+y)^2-xy-e^(3x+7y) ?

##### 2 Answers
Jun 24, 2016

$y ' = \frac{2 x + y - 3 {e}^{3 x + 4 y}}{4 {e}^{3 x + 4 y} - x - 2 y} .$

#### Explanation:

Given Eqn. is $- 1 = {\left(x + y\right)}^{2} - x y - {e}^{3 x + 7 y} ,$ or,
$x y + {e}^{3 x + 4 y} = {\left(x + y\right)}^{2} + 1.$

Diff. both sides,

$\left(x y\right) ' + \left\{{e}^{3 x + \setminus \textcolor{red}{7} y}\right\} ' = \left\{{\left(x + y\right)}^{2}\right\} ' + 0.$
$\therefore x y ' + y x ' + {e}^{3 x + 7 y} \left(3 x + 7 y\right) ' = 2 \left(x + y\right) \left(x + y\right) '$
$\therefore x y ' + y + {e}^{3 x + 7 y} \left\{\left(3 x\right) ' + \left(7 y\right) '\right\} = 2 \left(x + y\right) \left(x ' + y '\right)$
$\therefore x y ' + y + {e}^{3 x + 7 y} \left(3 + 7 y '\right) = 2 \left(x + y\right) \left(1 + y '\right) ,$ i.e.,
$x y ' + y + 3 {e}^{3 x + 4 y} + 7 y ' {e}^{3 x + 7 y} = 2 \left(x + y\right) + 2 y ' \left(x + y\right) .$
$\therefore x y ' + 7 y ' {e}^{3 x + 7 y} - 2 y ' \left(x + y\right) = 2 \left(x + y\right) - y - 3 {e}^{3 x + 4 y} .$
$\therefore y ' \left(x + 7 {e}^{3 x + 7 y} - 2 x - 2 y\right) = 2 x + 2 y - y - 3 {e}^{3 x + 4 y} ,$ or,
$y ' \left(7 {e}^{3 x + 7 y} - x - 2 y\right) = 2 x + y - 3 {e}^{3 x + 4 y} .$

Hence, $y ' = \frac{2 x + y - 3 {e}^{3 x + 7 y}}{7 {e}^{3 x + 7 y} - x - 2 y} .$

$y '$ can further be simplified, as below :-
For this, we write the given eqn. as ${e}^{3 x + 7 y} = {\left(x + y\right)}^{2} + 1 - x y = {x}^{2} + x y + {y}^{2} + 1 ,$ & submit the value of e^(3x+7y)) in $y '$ to give,

$y ' = \frac{2 x + y - 3 \left({x}^{2} + x y + {y}^{2} + 1\right)}{4 \left({x}^{2} + x y + {y}^{2} + 1\right) - x - 2 y} .$

Jun 24, 2016

$y ' = \frac{2 x + y - 3 \left({x}^{2} + {y}^{2} + x y + 1\right)}{7 \left({x}^{2} + {y}^{2} + x y + 1\right) - \left(2 y + x\right)} .$

#### Explanation:

We will use the given eqn. as ${e}^{3 x + 7 y} = {\left(x + y\right)}^{2} - x y + 1 = {x}^{2} + {y}^{2} + x y + 1 ,$ & diff. its both sides, to get,

${e}^{3 x + 7 y} \cdot \left(3 + 7 y '\right) = 2 x + 2 y y ' + x y ' + y .$
Subbing ${x}^{2} + {y}^{2} + x y + 1$ for ${e}^{3 x + 7 y}$ in $L . H . S . ,$

$3 \left({x}^{2} + {y}^{2} + x y + 1\right) + 7 y ' \left({x}^{2} + {y}^{2} + x y + 1\right) = 2 x + y + \left(2 y + x\right) y '$

$y ' \left\{7 \left({x}^{2} + {y}^{2} + x y + 1\right) - \left(2 y + x\right)\right\} = 2 x + y - 3 \left({x}^{2} + {y}^{2} + x y + 1\right)$

$y ' = \frac{2 x + y - 3 \left({x}^{2} + {y}^{2} + x y + 1\right)}{7 \left({x}^{2} + {y}^{2} + x y + 1\right) - \left(2 y + x\right)} .$

I find this soln. of mine very easier than the one I provided earlier!