How do you implicitly differentiate 2= ysinx-xcosy ?

$y ' = \frac{\cos y - y \cdot \cos x}{\sin x + x \cdot \sin y}$

Explanation:

Start with the given equation and differentiate with respect to x both sides of the equation

$2 = y \cdot \sin x - x \cdot \cos y$

$\frac{d}{\mathrm{dx}} \left(2\right) = \frac{d}{\mathrm{dx}} \left(y \cdot \sin x - x \cdot \cos y\right)$

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

$0 = y \cdot \cos x + \sin x \cdot y ' - \left[x \left(- \sin y \cdot y '\right) + \cos y \cdot 1\right]$

$0 = y \cdot \cos x + \sin x \cdot y ' + x \cdot \sin y \cdot y ' - \cos y$

transpose terms without the $y '$

$\cos y - y \cdot \cos x = \sin x \cdot y ' + x \cdot \sin y \cdot y '$

by symmetric property of equality we have

$\sin x \cdot y ' + x \cdot \sin y \cdot y ' = \cos y - y \cdot \cos x$

factor out $y '$

$\left(\sin x + x \cdot \sin y\right) \cdot y ' = \cos y - y \cdot \cos x$

Divide both sides of the equation by $\left(\sin x + x \cdot \sin y\right)$

$\frac{\left(\sin x + x \cdot \sin y\right) \cdot y '}{\sin x + x \cdot \sin y} = \frac{\cos y - y \cdot \cos x}{\sin x + x \cdot \sin y}$

$\frac{\cancel{\sin x + x \cdot \sin y} \cdot y '}{\cancel{\sin x + x \cdot \sin y}} = \frac{\cos y - y \cdot \cos x}{\sin x + x \cdot \sin y}$

$y ' = \frac{\cos y - y \cdot \cos x}{\sin x + x \cdot \sin y}$

God bless....I hope the explanation is useful.