If `y^2 = p ( x ),` a polynomial of degree 3, then `2 d/(dx) (y^3 (d^2y)/(dx^2))` equals :

Options

A. `p'''(x) + p '(x)`
B. `P''(x).P'''(x)`
C. `P(x)P'''(x)`
D. a constant

Answer

B

Explanation

Since , ` y^(2) = P(x)`
On differentiating both sides , we get
` 2yy_(1) = P'(x)`
Again , differentiating , we get
`2yy_(2) + 2y_(1)^(2) = P''(x) `
` rArr 2y^(3) y_(2) + 2y^(2) y_(1)^(2) = y^(2) P''(x) `
`rArr 2y^(3) y_(2) = y^(2) P''(x) - 2(yy_(1))^(2)`
`rArr 2y^(3) y_(2) = P(x). P ''(x) - ({P'(x)}^(2))/(2)`
Again differentiating , we get
`2(d)/(dx) (y^(3) y_(2)) = P'(x) . P''(x) + P(x).P'''(x)`
` - (2P'(x).P''(x))/(2)`
` rArr 2(d)/(dx) (y^(3) y_(2))= P(x).P'''(x)`
`rArr 2(d)/(dx) (y^(3) . (d^(2) y)/(dx^(2))) = P (x) . P '''(x)`