Question

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

Answer 1

`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)`