How do you find #dy/dx# by implicit differentiation of #y=sin(xy)#?

Question

Ezekiel Alexander · Accepted Answer

# dy/dx={ycos(xy)}/ {1-xcos(xy)},#

,OR,

#dy/dx={y^2sqrt(1-y^2)}/{y-sqrt(1-y^2)arc siny}.# #y=sin(xy).# #:. dy/dx," using the Chain Rule,"# #=d/dx(sin(xy))={cos(xy)}{d/dx(xy)}," &, using the Product Rule,"# #={x*d/dx(y)+y*d/dx(x)}cos(xy),# #:. dy/dx=xcos(xy)dy/dx+ycos(xy),# #rArr {1-xcos(xy)}dy/dx=ycos(xy).# #:. dy/dx={ycos(xy)}/ {1-xcos(xy)}.# Otherwise, #y=sin(xy) rArr arc siny=xy, or, x=(arc siny)/y.# Hence, diff.ing both sides w.r.t. #y,# we have, by the Quotient Rule, #dx/dy={y*d/dy(arc siny)-(arc siny)*d/dy(y)}/y^2,# #={y*(1/sqrt(1-y^2))-(arc siny)*1}/y^2,# #={y-sqrt(1-y^2)arc siny}/{y^2*sqrt(1-y^2)},# Therefore, #dy/dx={y^2sqrt(1-y^2)}/{y-sqrt(1-y^2)arc siny}.# I leave it to the Questioner to show that both Answers tally with each other. Enjoy Maths.!

Adam Adkins · Answer

undefined To find dy/dx by implicit differentiation of y = sin(xy), apply the chain rule and product rule. First, differentiate both sides of the equation with respect to x. The derivative of sin(xy) with respect to x is cos(xy) * (y + x * dy/dx). Then solve for dy/dx. So, dy/dx = (cos(xy) * y + cos(xy) * x * dy/dx) / (cos(xy) * x - sin(xy)). Finally, isolate dy/dx to get the result.