Question

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

Answer 1

To differentiate implicitly, simply differentiate each term with respect to `x`, then rearrange for `dy/dx`.

Answer:`dy/dx=(3x^2y+1)/(8y-x^3-1)`

Explanation:

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

I will take it piece by piece for the sake of clarity.

`d/(dx)(4y^2)=8ydy/dx`

This is from applying the chain rule.

`d/(dx)(x^3y)=x^3dy/dx+3x^2y`

This is from applying the product rule.

`d/(dx)(y)=dy/dx`

`d/(dx)(x)=1`

Putting that all together we get:

`8ydy/dx=x^3dy/dx+3x^2y+dy/dx+1`

Rearrange:

`dy/dx(8y-x^3-1)=3x^2y+1`

`dy/dx=(3x^2y+1)/(8y-x^3-1)`

Sometimes it is also possible to find `y` in terms of `x` in the original equation and sub back to get `dy/dx` only in terms of `x`.