Question

How do you find `(dy)/(dx)` given `2xy+y^2=x+y`?

Answer 1

`\frac{1-2y}{2x+2y-1}`

Explanation:

`2xy+y^2 = x+y`
`\frac{d}{dx}(2xy+y^{2})=\frac{d}{dx}(x+y)`
`2\frac{d}{dx}(xy)+2y\frac{d}{dx}=1+\frac{d}{dx}`

Use product rule and finish it off to get:

`2(x\frac{d}{dx}+y)+2y\frac{d}{dx}=1+\frac{d}{dx}`

to make it clearer, let `\frac{d}{dx}=y'`

so you get:
`2(xy'+y)+2yy'=1+y'`

Rearrange to make `y'` the factor to get:

`y'=\frac{1-2y}{2x+2y-1}`

bring back `y'=\frac{d}{dx}`

so you get:
`\frac{d}{dx}: \frac{1-2y}{2x+2y-1}`

Answer 2

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

Explanation:

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

`"noting that "d/dx(y)=dy/dx" and "d/dx(y^2)=2ydy/dx`

`"differentiate "xy" using the "color(blue)"product rule"`

`2xdy/dx+2y+2ydy/dx=1+dy/dx`

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

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