How do you implicitly differentiate -1=(x-y)sinx+y?

Nov 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin x + x \cos x - y \cos x}{\sin x - 1}$

Explanation:

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

$- 1 = x \sin x - y \sin x + y$

$0 = \sin x + x \cos x - y \cos x - \sin x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}}$

$\sin x \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = \sin x + x \cos x - y \cos x$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin x + x \cos x - y \cos x}{\sin x - 1}$