Question

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

Answer 1

`dy/dx=(e^(xy)-(x-y^2)e^(xy)y)/(1+2y-(x-y^2)e^(xy)x)`

Explanation:

`y=(x-y^2)e^(xy)`

We could proceed using the product rule. In implicit differentiation, we must also remember to use the chain rule when differentiating a `y` since `y` is a function of `x`. So:

`y=(x-y^2)e^(xy)`

`dy/dx=d/dx(x-y^2)*e^(xy)+(x-y^2)d/dxe^(xy)`

`dy/dx=(1-2ydy/dx)e^(xy)+(x-y^2)*e^(xy)*d/dx{xy}`

`dy/dx=(1-2ydy/dx)e^(xy)+(x-y^2)*e^(xy)(y+xdy/dx)`

Now gather all terms with `dy/dx` on the left side:

`dy/dx+2ydy/dx-(x-y^2)e^(xy)xdy/dx=e^(xy)+(x-y^2)e^(xy)y`

`dy/dx(1+2y-(x-y^2)e^(xy)x)=e^(xy)+(x-y^2)e^(xy)y`

`->dy/dx=(e^(xy)-(x-y^2)e^(xy)y)/(1+2y-(x-y^2)e^(xy)x)`

Answer 2

`dy/dx=(xye^(xy)-y^3e^(xy)+e^(xy))/(1-x^2e^(xy)+xy^2e^(xy)+2ye^(xy))`

Explanation:

`"differentiate using the "color(blue)"product/chain rules"`

`"given "y=f(x)g(x)" then"`

`dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`f(x)=x-y^2rarrf'(x)=1-2ydy/dx`

`g(x)=e^(xy)rArrg'(x)=e^(xy)xxd/dx(xy)`

`color(white)(xxxxxxxxxxxxx)=e^(xy)(xdy/dx+y)`

`color(white)(xxxxxxxxxxxxx)=xe^(xy)dy/dx+ye^(xy)`

`(x-y^2)(xe^(xy)dy/dx+ye^(xy)))+e^(xy)(1-2ydy/dx)`

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

`y=(x-y^2)e^(xy)larrcolor(blue)"original question"`

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

`rArrdy/dx(1-x^2e^(xy)+xy^2e^(xy)+2ye^(xy))=xye^(xy)-y^3e^(xy)+e^(xy)`

`rArrdy/dx=(xye^(xy)-y^3e^(xy)+e^(xy))/(1-x^2e^(xy)+xy^2e^(xy)+2ye^(xy))`