Question

How do you implicitly differentiate `y= y(x-y)^2 + e^(x y)`?

Answer 1

`dy/dx=[y{e^(xy)+2(x-y)}]/{1-(x-y)(x-3y)-xe^(xy)}.`

Explanation:

`y=y(x-y)^2+e^(xy) rArr d/dx(y)=d/dx{y(x-y)^2+e^(xy)}.`

`:. dy/dx=d/dx{y(x-y)^2}+d/dx{e^(xy)}.....(star).`

To differentiate the R.H.S., we use the Product Rule and the

Chain Rule :

`d/dx{y(x-y)^2}=yd/dx(x-y)^2+(x-y)^2d/dx(y),`

`=y{2(x-y)}d/dx(x-y)+(x-y)^2dy/dx,`

`=2y(x-y){d/dx(x)-d/dx(y)}+(x-y)^2dy/dx,`

`=2y(x-y){1-dy/dx}+(x-y)^2dy/dx,`

`=2y(x-y)+(x-y)^2dy/dx-2y(x-y)dy/dx,`

`=2y(x-y)+(x-y){(x-y)-2y}dy/dx,`

`=2y(x-y)+(x-y)(x-3y)dy/dx.............(1).`

Similarly, `d/dx{e^(xy)}=e^(xy)d/dx{xy},`

`=e^(xy){xd/dx(y)+yd/dx(x)},`

`=e^(xy){xdy/dx+y}=xe^(xy)dy/dx+ye^(xy).............(2).`

Altogether, by `(star), (1) and (2),`

`dy/dx=2y(x-y)+(x-y)(x-3y)dy/dx+xe^(xy)dy/dx+ye^(xy)`

`:. {1-(x-y)(x-3y)-xe^(xy)}dy/dx=2y(x-y)+ye^(xy)`

`rArr dy/dx=[y{e^(xy)+2(x-y)}]/{1-(x-y)(x-3y)-xe^(xy)}.`

Enjoy Maths.!