With the usual notation, \frac{d^2x}{dy^2} is :

Correct Answer is : −(d2ydx2) (dydx)−2

(\frac{d^2y}{dx^2}) (\frac{dy}{dx})^{-2}

-(\frac{d^2y}{dx^2})^{-1} (\frac{dy}{dx})^{-2}

-(\frac{d^2y}{dx^2}) (\frac{dy}{dx})^{-2}

NIMCET 2015 Question Paper with solutions

Question : 41 of 120

Marks: +1, -0

With the usual notation,

{d^2x}/{dy^2}

is :

Solution:

Since

{dx}/{dy}=1/{dy}/{dx}=({dy}/{dx})^{-1}

⇒{d}/{dy}({dx}/{dy})={d}/{dx}({dy}/{dx})^{-1}{dy}/{dx}

⇒{d^{2}x}/{dy^{2}}=-({d^{2}y}/{dx^{2}})({dy}/{dx})^{-2}({dy}/{dx})

⇒{d^{2}x}/{dy^{2}}=-({d^{2}y}/{dx^{2}})({dy}/{dx})^{-3}

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