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!
