Question

What is the value of the implicit derivative of `y=y^2x^2` at `x=3`?

Answer 1

`0,-2/27`

Explanation:

`d/dx[y=y^2x^2]`

`dy/dx=x^2d/dx[y^2]+y^2d/dx[x^2]`

`dy/dx=2x^2ydy/dx+2xy^2`

`dy/dx-2x^2ydy/dx=2xy^2`

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

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

Now that we have the implicit derivative, we must find the point or points when `x=3`, so we can plug them in to find the value(s) of the implicit derivative.

We know that `x=3`, so:

`y=y^2x^2`
`y=9y^2`

With this:

`0=9y^2-y`
`0=y(9y-1)`

`y=0,1/9`

We now have to two points `(3,0)` and `(3,1/9)`.

`(3,0)rarr`

`dy/dx=(2(3)(0^2))/(1-2(3^2)(0))=0/1=0`

`(3,1/9)rarr`

`dy/dx=(2(3)(1/9)^2)/(1-2(3^2)(1/9))=(2/27)/(-1)=-2/27`