Question

How do you implicitly differentiate `-1=-y^2x-2xy-ye^y`?

Answer 1

`y'=(-y^2-2y)/(2xy+2x+e^y+ye^y)`

Explanation:

So firstly differentiate all parts individually

`-1=y^2x-2xy-ye^y`
`-1 => 0`
`-y^2x =>-2yxdy/dx+y^2`
`-2xy=>-2dy/dx-2y`
`-ye^y=>-e^ydy/dx-ye^ydy/dx`

(this is done using mostly the product rule i.e.
`y=f(x)g(x)`
`y'=f'(x)g(x)+f(x)g'(x)`)

leaving us with,

`0=-2yx dy/dx-y^2-2y-2xdy/dx-e^ydy/dx-ye^ydy/dx`

`2yx dy/dx +2xdy/dx+e^ydy/dx+ye^ydy/dx=-y^2-2y`

`dy/dx(2yx+2x+e^y+ye^y)=-y^2-2y`

`dy/dx=(-y^2-2y)/(2yx+2x+e^y+ye^y)`