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

##### 1 Answer
Sep 27, 2016

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

#### Explanation:

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

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

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

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

$0 = y {e}^{x y} + x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \sin y + {y}^{3} + 3 x {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Collecting all similar monomials including $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$0 = x {e}^{x y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \sin y + 3 x {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y {e}^{x y} + {y}^{3}$

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

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

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