Question

How do you differentiate `x * y + 2x + 3x^2 = 4`?

Answer 1

`d/dx[xy+2x+3x^2=4]` results in `dy/dx=3x-4-6/x`

Explanation:

Using Implicit Differentiation we have

`d/dx[xy+2x+3x^2=4]`
`d/dx[xy]+d/dx[2x]+d/dx[3x^2]=0`
`x*d/dx[y]+y*d/dx[x]+2*d/dx[x]+3*d/dx[x^2]=0`
`x*y'+y*1+2*1+3*(2x)=0`
`x*y'+y=-6x-2`
`y'=(-6x-2)/x-y`

Solving for y in the original problem, we have

`x*y+2x+3x^2=4`
`x*y=4-2x-3x^2`
`y=(4-2x-3x^2)/x`

Now, plug this value for `y` into the equation for `y'` above

`y'=(-6x-2)/x-y`
`y'=(-6x-2)/x-(4-2x-3x^2)/x`
`y'=(-6x-2-4+2x+3x^2)/x`
`y'=(-4x-6+3x^2)/x`
`y'=(3x^2-4x-6)/x`
`y'=3x-4-6/x`