Question

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

Answer 1

`2sinxcosy = 1`

`=>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!