dx/dy = ≤ft(dy/dx)^-1 Differentiating with respect to y : d^2 x/dy^2 = d/dy ≤ft[ ≤ft(dy/dx)^-1 ] = d/dx ≤ft[ ≤ft(dy/dx)^-1 ] dx/dy Using power rule on first part: d^2 x/dy^2 = ≤ft[ -1 ≤ft(dy/dx)^-2 × d^2 y/dx^2 ] × ≤ft(dy/dx)^-1 d^2 x/dy^2 = -≤ft(d^2 y/dx^2) ≤ft(dy/dx)^-3

d^2 x/dy^2 equals: - Mathematics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

hardMCQFIITJEE AIEEE 2011 (Set Q)2026Mathematics

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\frac{d ^{2} x}{d y ^{2}} equals:

A $- (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 3}$
B $(\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 2}$
C $- (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 2}$
D $(\frac{d ^{2} y}{d x ^{2}})^{- 1}$

Solution & Step-by-step Explanation

\frac{d x}{d y} = (\frac{d y}{d x})^{- 1} Differentiating with respect to y : \frac{d ^{2} x}{d y ^{2}} = \frac{d}{d y} [(\frac{d y}{d x})^{- 1}] = \frac{d}{d x} [(\frac{d y}{d x})^{- 1}] \frac{d x}{d y} Using power rule on first part: \frac{d ^{2} x}{d y ^{2}} = [- 1 (\frac{d y}{d x})^{- 2} \cdot \frac{d ^{2} y}{d x ^{2}}] \cdot (\frac{d y}{d x})^{- 1} \frac{d ^{2} x}{d y ^{2}} = - (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 3}

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\frac{d ^{2} x}{d y ^{2}} equals:

- (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 3}

(\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 2}

- (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 2}

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

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