# Question #54070

May 11, 2017

${y}^{2} = P$

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

First derivative wrt $x$:

$2 y {y}_{x} = {P}_{x}$

Second derivative wrt $x$:

$2 {\left({y}_{x}\right)}^{2} + 2 y y {\setminus}_{x x} = P {\setminus}_{x x}$

Third derivative wrt $x$:

$4 {y}_{x} {y}_{x x} + 2 {y}_{x} y {\setminus}_{x x} + 2 y y {\setminus}_{x x x} = P {\setminus}_{x x x}$

$\implies 6 {y}_{x} {y}_{x x} + 2 y y {\setminus}_{x x x} = P {\setminus}_{x x x}$

$\implies P {\setminus}_{x x x} = \textcolor{red}{2 \left(3 {y}_{x} {y}_{x x} + y y {\setminus}_{x x x}\right)}$

Now we have:

$2 \frac{d}{\mathrm{dx}} \left({y}^{3} {y}_{x x}\right) = 3 {y}^{2} {y}_{x} {y}_{x x} + {y}^{3} y {\setminus}_{x x x}$

$= {y}^{2} \cdot \textcolor{red}{2 \left(3 {y}_{x} {y}_{x x} + y y {\setminus}_{x x x}\right)}$

$= P \cdot {P}_{x x x}$