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

Sep 30, 2017

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

#### Explanation:

For this, we take derivatives of all terms with respect to $x$. When differentiating a $y$ term, we must note the presence of $\frac{\mathrm{dy}}{\mathrm{dx}}$ (often written as $y '$ for convenience). After that is done, we solve for $y '$:

$0 = 2 \left(x + y\right) \left(1 + 1 \cdot y '\right) - 2 y y ' - 1 + {e}^{3 x + {y}^{2}} \cdot \left(3 + 2 y y '\right)$

$0 = \left(2 x + 2 y\right) \left(1 + y '\right) - 2 y y ' - 1 + 3 {e}^{3 x + {y}^{2}} + 2 y y ' {e}^{3 x + {y}^{2}}$

$0 = 2 x + 2 x y ' + 2 y + 2 y y ' - 2 y y ' - 1 + 3 {e}^{3 x + {y}^{2}} + 2 y y ' {e}^{3 x + {y}^{2}}$

$0 = 2 x + 2 x y ' + 2 y - 1 + 3 {e}^{3 x + {y}^{2}} + 2 y y ' {e}^{3 x + {y}^{2}}$

$- 2 x - 2 y + 1 - 3 {e}^{3 x + {y}^{2}} = 2 x y ' + 2 y y ' {e}^{3 x + {y}^{2}}$

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

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