Question

How do you find the second derivative of `x^2y^2=1`?

Answer 1

You can calculate a derivative of a function, but `x2y^2-1` is not a function, it's an equality...

Maybe you're looking for derivative `f(x,y) = x^2y^2-1` ? But for which variable ? `x` or `y` ?

Answer 2

First we need to find the first derivative of the function.
We will do this using implicit differentiation. With this
particular function we will use the product rule.

`2xy^2+x^2(2y)(dy/dx)=0`

Now subtract `2xy^2` from both sides

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

Divide both sides by `x^2(2y)`

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

Which simplifies to

`dy/dx=-y/x`

Now for the second derivative we will use the quotient rule

`(d^2 y)/dx^2=[(x(-1)dy/dx)-(1(-y))]/x^2`

`(d^2 y)/dx^2=(-xdy/dx+y)/x^2=(y-xdy/dx)/x^2`

plug `dy/dx` into the right hand side

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

`(d^2 y)/dx^2=(2y)/x^2`