from the given
#4*cos xy=sin 2x#
if #x=1# then
#4*cos y=sin 2# and
#y=cos^(-1)(0.25*sin 2)#
#y=1.341467106#
therefore the point of tangency is #(1, 1.341467106)# and not #(1, 1)#
by Implicit differentiation
#-4*sin(xy)*(xy'+y)=2*cos (2x)#
solving for #y'#
#y'=(-cos 2x)/(2x*sin (xy))-y/x#
#y'=(-cos (2*1))/(2*1*sin (1*cos^(-1)(0.25*sin 2)))-(cos^(-1)(0.25*sin 2))/1#
#y'=-1.127799682# this is the slope
Using point slope form
#y-y_1=m*(x-x_1)#
with #x_1=1# and #y_1=1.341467106#
Equation of the tangent line at #(1, 1.341467106)#
#y-1.341467106=-1.127799682*(x-1)#
#y=-1.127799682*x+1.127799682+1.341467106#
#y=-1.127799682*x+2.469266788# graph{4*cos (xy)=sin (2x) [-10, 10, -5, 5]}
graph{y=-1.127799682*x+2.469266788[-10,10,-5,5]}
