Question
If $${y^2} = p\left( x \right)$$ is a polynomial of degree $$3,$$ then what is $$2\frac{d}{{dx}}\left[ {{y^3}\frac{{{d^2}y}}{{d{x^2}}}} \right]$$ equal to ?
A.
$$p'\left( x \right)p'''\left( x \right)$$
B.
$$p''\left( x \right)p'''\left( x \right)$$
C.
$$p\left( x \right)p'''\left( x \right)$$
D.
A constant
Answer :
$$p\left( x \right)p'''\left( x \right)$$
Solution :
Given that $${y^2} = p\left( x \right)$$
Differentiating
$$\eqalign{
& \Rightarrow 2y{y_1} = p'\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{here }}{y_1} = \frac{{dy}}{{dx}}} \right] \cr
& \Rightarrow 2{y_1} = \frac{{p'\left( x \right)}}{y} \cr} $$
Differentiating again,
$$\eqalign{
& \Rightarrow 2{y_2} = \frac{{yp''\left( x \right) - p'\left( x \right){y_1}}}{{{y^2}}},\,\,\,\,\,\,\,\,\,\,\left[ {{\text{here }}{y_2} = \frac{{{d^2}y}}{{d{x^2}}}} \right] \cr
& \Rightarrow 2{y_2} = \frac{{yp''\left( x \right) - \frac{{p'\left( x \right).p'\left( x \right)}}{{2y}}}}{{{y^2}}} \cr
& \Rightarrow 2{y_2} = \frac{{2{y^2}p''\left( x \right) - {{\left( {p'\left( x \right)} \right)}^2}}}{{{2y^3}}} \cr
& \Rightarrow 2{y^3}{y_2} = \frac{1}{2}\left[ {2{y^2}p''\left( x \right) - {{\left( {p'\left( x \right)} \right)}^2}} \right] \cr
& \Rightarrow 2{y^3}{y_2} = \frac{1}{2}\left[ {2p\left( x \right)p''\left( x \right) - {{\left( {p'\left( x \right)} \right)}^2}} \right] \cr
& \Rightarrow 2\frac{d}{{dx}}\left( {{y^3}{y_2}} \right) = \frac{1}{2}\left[ {2p'\left( x \right) + 2p\left( x \right)p'''\left( x \right) - 2p'\left( x \right)p''\left( x \right)} \right] \cr
& \Rightarrow 2\frac{d}{{dx}}\left( {{y^3}{y_2}} \right) = p\left( x \right)p'''\left( x \right) \cr} $$