Question

Question #54070

Answer 1

`y^2 = P`

What follows is mostly chain and product rules with implicit differentiation.

First derivative wrt `x`:

`2 y y_x = P_x`

Second derivative wrt `x`:

`2 (y_x)^2 + 2 y y_( x x ) = P_(x x)`

Third derivative wrt `x`:

`4 y_x y _(x x) + 2 y_x y_( x x ) + 2 y y_( x x x )= P_(x x x)`

`implies 6 y_x y _(x x) + 2 y y_( x x x )= P_(x x x)`

`implies P_(x x x) = color(red)(2 ( 3 y_x y _(x x) + y y_( x x x )) )`

Now we have:

`2d/dx (y^3 y_(x x) ) = 3 y^2 y_x y _(x x) + y^3 y_(x x x)`

`=y^2 * color(red)(2 (3 y_x y _(x x) + y y_(x x x) ))`

`= P * P_(x x x)`