# How do you find dy/dx by implicit differentiation of 2sinxcosy=1?

Dec 12, 2016

$2 \sin x \cos y = 1$

$\implies \sin x \cos y = \frac{1}{2}$

$\implies \frac{d}{\mathrm{dx}} \left(\sin x \cos y\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{2}\right)$

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

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

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

Hopefully this helps!