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

Jul 21, 2016

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

#### Explanation:

When we implicitly differentiate a function $f \left(y\right)$, we first differentiate $f \left(y\right)$ with respect to $y$ and then multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$. This comes from chain rule as $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \times \frac{\mathrm{dy}}{\mathrm{dx}}$.

Now using product rule for differentiating $1 = - {y}^{2} x - 2 x y - y {e}^{x y}$, we get

$0 = - \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}} \times x + {y}^{2}\right) - 2 \left(1 \times y + x \times \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left(\frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x y} + y \times {e}^{x y} \left(1 \times y + x \times \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)$ or

$\left(2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2}\right) + 2 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(\frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x y} + y {e}^{x y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) = 0$ or

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

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