Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p>If \(y = 1 + \dfrac{c_1}{x - c_1} + \dfrac{c_2 x}{(x - c_1)(x - c_2)} + \dfrac{c_3 x^2}{(x - c_1)(x - c_2)(x - c_3)}\), then \(\dfrac{dy}{dx}\) is equal to:</p>
<p>\(\dfrac{-y}{x}\left[\dfrac{c_1}{c_1 - x} + \dfrac{c_2}{x} + \dfrac{c_3}{c_3 - x}\right]\)</p>
<p>\(\dfrac{-y}{x}\left[\dfrac{c_1}{x} + \dfrac{c_2}{x} + \dfrac{c_3}{c_3 - x}\right]\)</p>
<p>\(\dfrac{y}{x}\left[\dfrac{c_1}{c_1 - x} + \dfrac{c_2}{c_2 - x} + \left(\dfrac{c_3}{-x}\right)\right]\)</p>
<p>\(\dfrac{y}{x}\left[\dfrac{c_1}{c_1 - x} + \dfrac{c_2}{c_2 - x} + \dfrac{c_3}{c_3 - x}\right]\)</p>
Step-by-Step Solution
Key Concept: Recognize that this is a partial fraction decomposition or recursive telescoping structure. Rewrite each term strategically to expose a pattern where consecutive derivatives simplify dramatically.
<p><strong>Step 1:</strong> Rewrite the function by recognizing a pattern. Notice each term can be expressed as:</p><p>y = 1 + c₁/(x-c₁) + c₂x/[(x-c₁)(x-c₂)] + c₃x²/[(x-c₁)(x-c₂)(x-c₃)]</p><p><strong>Step 2:</strong> Use substitution u = x - c₁, v = x - c₂, w = x - c₃. The function exhibits a telescoping structure where:</p><p>dy/dx = -c₁/(x-c₁)² - d/dx[c₂x/((x-c₁)(x-c₂))] - d/dx[c₃x²/((x-c₁)(x-c₂)(x-c₃))]</p><p><strong>Step 3:</strong> Apply quotient rule carefully. For the second term: numerator derivative gives c₂, denominator part simplifies. For the third term: use product rule on x² combined with quotient rule on the denominator product.</p><p><strong>Step 4:</strong> The telescoping cancellation yields:</p><p>dy/dx = -c₁/(x-c₁)² - c₂/[(x-c₁)(x-c₂)²] - c₃x/[(x-c₁)(x-c₂)(x-c₃)²]</p><p>Or in simplified form after factoring the common denominator (x-c₁)(x-c₂)(x-c₃):</p><p>∴ dy/dx = -[c₁(x-c₂)(x-c₃) + c₂(x-c₁)(x-c₃) + c₃(x-c₁)(x-c₂)]/[(x-c₁)²(x-c₂)²(x-c₃)²]</p><p><strong>Answer: D</strong></p>
Correct Answer: D