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

Question

Charles Anctil · Accepted Answer

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)#

Axel Alvarez · Answer

undefined To find the second derivative of $x^2y^2$, you'll need to use the product rule and chain rule twice. Here's the step-by-step process:

1. Find the first derivative using the product rule:
   $$ \frac{d}{dx}(x^2y^2) = 2x \cdot y^2 + x^2 \cdot 2y \frac{dy}{dx} $$

2. Simplify the expression:
   $$ \frac{d}{dx}(x^2y^2) = 2xy^2 + 2x^2y\frac{dy}{dx} $$

3. Now, differentiate the expression obtained in step 2 with respect to $x$ to find the second derivative:
   $$ \frac{d^2}{dx^2}(x^2y^2) = 2y^2 + 4xy\frac{dy}{dx} + 2x^2\frac{d^2y}{dx^2} $$

This is the second derivative of $x^2y^2$.