\[\begin{array}{1 1}(1)1&(2)2\\(3)3&(4)6\end{array}\]

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$\sqrt {1+(\large\frac{dy}{dx})^{1/3} } =\frac{d^2 y}{dx^2}$

squaring $1+ (\large\frac{dy}{dx} )^{1/3}=( \frac{d^2 y}{dx^2})^2$

$(\large\frac{dy}{dx} )^{1/3}=( \frac{d^2 y}{dx^2})^2$$-1$

Raising both sides of the power 3.

$\large\frac{dy}{dx} = \bigg[ \bigg( \large\frac{d^2y)^2}{dx^2} \bigg)-1\bigg]^3$

Order - 2

degree -6

Hence 4 is the correct answer.

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