Question

If $ y = f(x), \, p = \frac{dy}{dx}, \, q = \frac{d^2 y}{dx^2}, \text{ then } \frac{d^2 x}{dy^2} \text{ is equal to } $

• A. \( \frac{-q}{p^2} \)
• B. \( \frac{q}{p} \)
• C. \( \frac{-q}{p^3} \)
• D. \( \frac{q}{p^2} \)

Hint:

When differentiating with respect to \( y \) using the chain rule, remember to also differentiate the terms involving \( x \) and then apply the appropriate inverse relationships for derivatives of \( x \) and \( y \).

Answer 1

The Correct Option is C

We are given the following relationships: \[ p = \frac{dy}{dx}, \quad q = \frac{d^2 y}{dx^2} \] We need to find the expression for \( \frac{d^2 x}{dy^2} \).
Step 1: Use the chain rule
First, we use the chain rule to express \( \frac{d^2 x}{dy^2} \) in terms of \( p \) and \( q \). The inverse of \( p = \frac{dy}{dx} \) gives us \( \frac{dx}{dy} = \frac{1}{p} \). To compute \( \frac{d^2 x}{dy^2} \), we differentiate \( \frac{dx}{dy} \) with respect to \( y \): \[ \frac{d^2 x}{dy^2} = \frac{d}{dy} \left( \frac{1}{p} \right) \] 
Step 2: Differentiate with respect to \( y \)
Using the chain rule again: \[ \frac{d^2 x}{dy^2} = \frac{-1}{p^2} \frac{dp}{dy} \] Now, we need to compute \( \frac{dp}{dy} \). We use the chain rule again, noting that \( p = \frac{dy}{dx} \) and \( q = \frac{d^2 y}{dx^2} \): \[ \frac{dp}{dy} = \frac{dp}{dx} \cdot \frac{dx}{dy} = q \cdot \frac{1}{p} \] 
Step 3: Final expression
Substituting into the expression for \( \frac{d^2 x}{dy^2} \): \[ \frac{d^2 x}{dy^2} = \frac{-1}{p^2} \cdot \frac{q}{p} = \frac{-q}{p^3} \] Thus, the correct answer is: \[ \frac{-q}{p^3} \]