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

Mar 24, 2015

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:

$\frac{d}{\mathrm{dx}} \left(F S T\right) = F ' S T + F S ' T + F S T '$

(I think using $\frac{d}{\mathrm{dx}}$ instead of 'prime' makes this harder to read.)

$\frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{2}\right) = 2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \left({x}^{2} {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(\textcolor{red}{2 x {y}^{2}} + \textcolor{b l u e}{2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}}\right)$

=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]

$= 2 {y}^{2} + 8 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {x}^{2} {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 2 {x}^{2} y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$