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Options

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

Answer

C

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
`2d/(dx) (y^(3)y_(2))=P'(x)*P''(x) + P(x)*P'''(x) -(2P'(x)*P''(x))/2`
` rArr" "2d/(dx) (y^(3)y_(2))=P(x)*P'''(x) `
` rArr" "2d/(dx) (y^(3)*(d^(2)y)/(dx^(2)))=P(x)*P'''(x) `