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 this as a telescoping partial fraction structure where successive terms contain increasing powers of x in numerators and cumulative factors in denominators. Differentiate by treating each fraction separately and notice the derivative produces a pattern that cancels internal terms.
<p><strong>Step 1:</strong> Rewrite the given expression by combining over a common denominator:</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> Recognize this has a telescoping structure. Rewrite systematically:</p><p>y = (x - c₁)/(x - c₁) + c₁/(x - c₁) + c₂x/[(x - c₁)(x - c₂)] + c₃x²/[(x - c₁)(x - c₂)(x - c₃)]</p><p><strong>Step 3:</strong> Group the first two terms: y = x/(x - c₁) + c₂x/[(x - c₁)(x - c₂)] + c₃x²/[(x - c₁)(x - c₂)(x - c₃)]</p><p><strong>Step 4:</strong> Continue telescoping: y = (x - c₂)/(x - c₂) + c₃x²/[(x - c₁)(x - c₂)(x - c₃)] = x/(x - c₂) + c₃x²/[(x - c₁)(x - c₂)(x - c₃)]</p><p><strong>Step 5:</strong> Final telescope gives: y = x/(x - c₃)</p><p><strong>Step 6:</strong> Differentiate: dy/dx = [(x - c₃)·1 - x·1]/(x - c₃)² = -c₃/(x - c₃)²</p><p>∴ Answer: D</p>
Correct Answer: D