Question #54070

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Answer 1 · 2017-05-11T20:26:13

#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)#