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

1 Answer
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:

#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)#