Limits, Continuity & Differentiability
Higher Order Derivatives
Grade 12
Question:
<p>\( \dfrac{d^2x}{dy^2} \) equals</p>
<p>\( -\left(\dfrac{d^2y}{dx^2}\right)^{-1}\left(\dfrac{dy}{dx}\right)^{-3} \)</p>
<p>\( \left(\dfrac{d^2y}{dx^2}\right)\left(\dfrac{dy}{dx}\right)^{-2} \)</p>
<p>\( -\left(\dfrac{d^2y}{dx^2}\right)\left(\dfrac{dy}{dx}\right)^{-3} \)</p>
<p>\( \left(\dfrac{d^2y}{dx^2}\right)^{-1} \)</p>
Step-by-Step Solution
Key Concept: The second derivative with respect to y requires recognizing that dx/dy = 1/(dy/dx), then differentiating this with respect to y (not x). Use the chain rule: d²x/dy² = d/dy[1/(dy/dx)] = -1/(dy/dx)² · d²y/dx².
<p><strong>Step 1:</strong> Start with the relationship between derivatives: if y = f(x), then dy/dx and dx/dy are reciprocals.</p><p><strong>Step 2:</strong> Therefore, dx/dy = 1/(dy/dx)</p><p><strong>Step 3:</strong> Differentiate both sides with respect to y using the chain rule:</p><p>d²x/dy² = d/dy[1/(dy/dx)] = d/dy[(dy/dx)⁻¹]</p><p><strong>Step 4:</strong> Apply chain rule: d²x/dy² = -(dy/dx)⁻² · d/dy(dy/dx)</p><p><strong>Step 5:</strong> Since d/dy(dy/dx) = (d²y/dx²)·(dx/dy) [chain rule with respect to y]</p><p><strong>Step 6:</strong> Substitute: d²x/dy² = -(dy/dx)⁻² · (d²y/dx²)·(1/(dy/dx))</p><p><strong>Step 7:</strong> Simplify: d²x/dy² = <strong>−(d²y/dx²)/(dy/dx)³</strong></p><p>∴ Answer: A</p>
Correct Answer: A