<p>If <math>y = (1 + x)(1 + x^2)(1 + x^4) \cdots (1 + x^{2^n})</math>, then the value of <math>\frac{dy}{dx}</math> at <math>x = 0</math> is</p>
Step-by-Step Solution
Key Concept: Use the algebraic identity (1-x)(1+x)(1+x²)(1+x⁴)···(1+x^(2^n)) = 1-x^(2^(n+1)) to simplify the product, then differentiate. Alternatively, recognize that at x=0, only the constant term of the expansion matters for dy/dx.
<p><strong>Step 1: Use the Algebraic Identity</strong></p><p>Multiply both sides by (1-x):</p><p>(1-x)·y = (1-x)(1+x)(1+x²)(1+x⁴)···(1+x^(2^n))</p><p>Using the identity (1-a)(1+a) = 1-a²:</p><p>(1-x)·y = (1-x²)(1+x²)(1+x⁴)···(1+x^(2^n))</p><p>Continuing this telescoping process:</p><p>(1-x)·y = 1 - x^(2^(n+1))</p><p><strong>Step 2: Expand y</strong></p><p>Therefore: y = (1 - x^(2^(n+1)))/(1-x)</p><p><strong>Step 3: Find dy/dx Using Quotient Rule</strong></p><p>Let u = 1 - x^(2^(n+1)) and v = 1-x</p><p>du/dx = -2^(n+1)·x^(2^(n+1)-1)</p><p>dv/dx = -1</p><p>dy/dx = [v(du/dx) - u(dv/dx)]/v²</p><p>dy/dx = [(1-x)·(-2^(n+1)·x^(2^(n+1)-1)) - (1-x^(2^(n+1)))·(-1)]/(1-x)²</p><p><strong>Step 4: Evaluate at x = 0</strong></p><p>At x = 0:</p><p>dy/dx|_(x=0) = [(1-0)·(-2^(n+1)·0) - (1-0)·(-1)]/(1-0)²</p><p>dy/dx|_(x=0) = [0 + 1]/1 = 1</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C