Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p>If \(x^m \cdot y^n = (x+y)^{m+n}\), then \(\dfrac{dy}{dx}\) is</p>
<p>\(\dfrac{y}{x}\)</p>
<p>\(\dfrac{x+y}{xy}\)</p>
<p>\(xy\)</p>
<p>\(\dfrac{x}{y}\)</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the equation x^m · y^n = (x+y)^(m+n), recognizing that both sides contain y as a function of x. The left side requires the product rule while the right side uses chain rule.
<p><strong>Step 1:</strong> Differentiate both sides with respect to x using implicit differentiation.</p><p>Left side: d/dx[x^m · y^n] = m·x^(m-1)·y^n + x^m·n·y^(n-1)·(dy/dx)</p><p><strong>Step 2:</strong> Right side: d/dx[(x+y)^(m+n)] = (m+n)(x+y)^(m+n-1)·(1 + dy/dx)</p><p><strong>Step 3:</strong> Set them equal: m·x^(m-1)·y^n + x^m·n·y^(n-1)·(dy/dx) = (m+n)(x+y)^(m+n-1)·(1 + dy/dx)</p><p><strong>Step 4:</strong> Expand right side: (m+n)(x+y)^(m+n-1) + (m+n)(x+y)^(m+n-1)·(dy/dx)</p><p><strong>Step 5:</strong> Collect dy/dx terms: [x^m·n·y^(n-1) - (m+n)(x+y)^(m+n-1)]·(dy/dx) = (m+n)(x+y)^(m+n-1) - m·x^(m-1)·y^n</p><p><strong>Step 6:</strong> From the original equation x^m·y^n = (x+y)^(m+n), substitute to simplify:</p><p>∴ dy/dx = -y(m·x + n·y)/(x(n·x + m·y)) or dy/dx = -y/x · (m+n·y/x)/(m·y/x+n)</p><p><strong>Simplified form: dy/dx = -y/x</strong> when m=n, or more generally <strong>dy/dx = y(my + nx)/(x(mx + ny))</strong></p>
Correct Answer: A