Q68 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { If } y=f(x), \quad p=\dfrac{d y}{d x} ; q=\dfrac{d^2 y}{d x^2} \text { then } \dfrac{d^2 x}{d y^2} \text { is equal to }$
- A$\dfrac{q}{p^2}$
- Correct answerB$-\dfrac{q}{p^3}$
- C$\dfrac{q}{p}$
- D$-\dfrac{q}{p^2}$
Solution
Given $p = \dfrac{dy}{dx}$, we have $\dfrac{dx}{dy} = \dfrac{1}{p} = p^{-1}$.
Differentiating both sides with respect to $y$ using the chain rule:
$\dfrac{d^2x}{dy^2} = \dfrac{d}{dy} \left( p^{-1} \right) = \dfrac{d}{dx} \left( p^{-1} \right) \cdot \dfrac{dx}{dy}$
$\dfrac{d^2x}{dy^2} = -p^{-2} \cdot \dfrac{dp}{dx} \cdot \dfrac{1}{p}$
Since $p = \dfrac{dy}{dx}$, then $\dfrac{dp}{dx} = \dfrac{d^2y}{dx^2} = q$.
Substituting these values:
$\dfrac{d^2x}{dy^2} = -\dfrac{1}{p^2} \cdot q \cdot \dfrac{1}{p} = -\dfrac{q}{p^3}$
Answer: $-\dfrac{q}{p^3}$
Differentiating both sides with respect to $y$ using the chain rule:
$\dfrac{d^2x}{dy^2} = \dfrac{d}{dy} \left( p^{-1} \right) = \dfrac{d}{dx} \left( p^{-1} \right) \cdot \dfrac{dx}{dy}$
$\dfrac{d^2x}{dy^2} = -p^{-2} \cdot \dfrac{dp}{dx} \cdot \dfrac{1}{p}$
Since $p = \dfrac{dy}{dx}$, then $\dfrac{dp}{dx} = \dfrac{d^2y}{dx^2} = q$.
Substituting these values:
$\dfrac{d^2x}{dy^2} = -\dfrac{1}{p^2} \cdot q \cdot \dfrac{1}{p} = -\dfrac{q}{p^3}$
Answer: $-\dfrac{q}{p^3}$
Hence, the correct answer is option: 'B'.