Differentiate both sides with respect to #x# and solve for #dy/dx.# Differentiating with respect to #x# means that every time we differentiate a term containing #y#, we must end up with an instance of #dy/dx:#
#d/dx1=d/dx(sin(2x)/y)#
#d/dx(1)=0#, so the left side becomes #0.#
Differentiate the right side using the quotient rule, #(f/g)'=(gf'-fg')/g^2#
#0=(y(2cos(2x))-dy/dxsin(2x))/y^2#
We want to solve for #dy/dx#. Multiply both sides by #y^2#. #0*y^2=0#, so the left side is still #0.#
#0=2ycos(2x)-dy/dxsin(2x)#
#dy/dxsin(2x)=2ycos(2x)#
#dy/dx=(2ycos(2x))/sin(2x)=2ycot(2x)# (as #cos(x)/sin(x)=cot(x)#
We'll keep this implicit solution containing #y# as we want the implicit derivative.
