How do you implicitly differentiate #2=ytanxy-xy #?

1 Answer
Aug 6, 2017

#dy/dx=(y-y^2sec^2(xy))/(xysec^2(xy)+tan(xy)-x)#

Explanation:

#"differentiate "color(blue)"implicitly with respect to x"#

#"differentiate "ytan(xy)" using the "color(blue)"product rule"#

#d/dx(ytan(xy))#

#=ysec^2(xy).(xdy/dx+y)+tan(xy).dy/dx#

#=xysec^2(xy)dy/dx+y^2sec^2(xy)+tan(xy)dy/dx#

#"differentiate "-xy" using the "color(blue)"product rule"#

#d/dx(-xy)=-xdy/dx-y#

#"putting the right side all together"#

#xysec^2(xy)dy/dx+y^2sec^2(xy)+tan(xy)dy/dx-xdy/dx-y#

#=dy/dx(xysec^2(xy)+tan(xy)-x)+y^2sec^2(xy)-y=0#

#rArrdy/dx=(y-y^2sec^2(xy))/(xysec^2(xy)+tan(xy)-x)#