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

##### 1 Answer

Dec 12, 2016

#=>sinxcosy = 1/2#

#=>d/dx(sinxcosy)= d/dx(1/2)#

#=> cosx(cosy) + sinx(-siny)(dy/dx) = 0#

#=>cosxcosy = sinxsiny(dy/dx)#

#=>(cosxcosy)/(sinxsiny) = dy/dx#

Hopefully this helps!