Question #027a6

1 Answer

#y=-1.127799682*x+2.469266788#

Explanation:

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]}