# How do you find the derivative of 5=3e^(xy)+x^2y+xy^2?

Oct 27, 2016

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

#### Explanation:

We need to use implicit differentiation and the product rule;
$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

$5 = 3 {e}^{x} y + {x}^{2} y + x {y}^{2}$

$\therefore 0 = 3 \left\{{e}^{x} \frac{d}{\mathrm{dx}} y + y \frac{d}{\mathrm{dx}} {e}^{x}\right\} + \left\{{x}^{2} \frac{d}{\mathrm{dx}} y + y \frac{d}{\mathrm{dx}} {x}^{2}\right\} + \left\{x \frac{d}{\mathrm{dy}} {y}^{2} + {y}^{2} \frac{d}{\mathrm{dx}} x\right\}$

$\therefore 0 = 3 \left\{{e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} + y {e}^{x}\right\} + \left\{{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + y \left(2 x\right)\right\} + \left\{x \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + {y}^{2}\right\}$

$\therefore 0 = 3 {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 y {e}^{x} + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2}$
$\therefore \left\{3 {e}^{x} + {x}^{2} + 2 x y\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(3 y {e}^{x} + 2 x y + {y}^{2}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 y {e}^{x} + 2 x y + {y}^{2}}{3 {e}^{x} + {x}^{2} + 2 x y}$