Question

How do you find the second derivative of `x^2y^2`?

Answer 1

If you mean to ask about second partial derivatives of a function of 2 variables, see Daniel's answer.

If you want to assume that `y` is an unknown function of `x`, then used implicit differentiation and the product rule.

We'll need the product rule for three factors:

`d/(dx)(FST)=F'ST+FS'T+FST'`

(I think using `d/(dx)` instead of 'prime' makes this harder to read.)

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

`d^2/(dx^2)(x^2y^2)=d/(dx)(color(red)(2xy^2)+color(blue)(2x^2y(dy)/(dx)))`

`=color(red)([2y^2+2x*2y(dy)/(dx)]) + color(blue)([4xy(dy)/(dx)+2x^2(dy)/(dx) (dy)/(dx)+2x^2y (d^2y)/dx^2]`

`=2y^2+8xy (dy)/(dx) + 2x^2((dy)/(dx))^2+2x^2y (d^2y)/(dx^2)`