# What is the implicit derivative of 1=xy-sinxy?

Mar 20, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left[1 - \cos \left(x y\right)\right]}{\left[\left(\cos \left(x y\right) - x\right)\right]}$

#### Explanation:

$1 = x y - \sin \left(x y\right)$

Differentiate both sides with respect to $x \Rightarrow$

$D \left(0\right) = D \left[x y - \sin \left(x y\right)\right]$

$0 = x . \frac{\mathrm{dy}}{\mathrm{dx}} + y - \left[\cos \left(x y\right) \times \left(x . \frac{\mathrm{dy}}{\mathrm{dx}} + y\right)\right]$

$0 = x . \frac{\mathrm{dy}}{\mathrm{dx}} + y - \left[\cos \left(x y\right) . x \frac{\mathrm{dy}}{\mathrm{dx}} + y \cos \left(x y\right)\right]$

$0 = x . \frac{\mathrm{dy}}{\mathrm{dx}} + y - \cos \left(x y\right) . \frac{\mathrm{dy}}{\mathrm{dx}} - y \cos \left(x y\right)$

$\cos \left(x y\right) . \frac{\mathrm{dy}}{\mathrm{dx}} - x . \frac{\mathrm{dy}}{\mathrm{dx}} = y - y \cos \left(x y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos \left(x y\right) - x\right) = y \left(1 - \cos \left(x y\right)\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left[1 - \cos \left(x y\right)\right]}{\left[\left(\cos \left(x y\right) - x\right)\right]}$