<p>If \(f(x) = \begin{vmatrix} 1 & 2x & 3x^2 \\ x & x^2 & x^3 \\ 0 & 2 & 6x \end{vmatrix}\) then \(f'(1)\) = ______</p>
Step-by-Step Solution
Key Concept: To find f'(1), we can use the property that the derivative of a determinant equals the sum of determinants obtained by differentiating one row at a time, then evaluate at x=1.
<p><strong>Step 1: Apply the determinant differentiation rule</strong></p><p>If f(x) = det(A(x)) where rows of A depend on x, then:</p><p>f'(x) = det(A'₁, A₂, A₃) + det(A₁, A'₂, A₃) + det(A₁, A₂, A'₃)</p><p>where A'ᵢ means the i-th row is differentiated.</p><p><strong>Step 2: Differentiate each row</strong></p><p>Row 1: (1, 2x, 3x²) → (0, 2, 6x)</p><p>Row 2: (x, x², x³) → (1, 2x, 3x²)</p><p>Row 3: (0, 2, 6x) → (0, 0, 6)</p><p><strong>Step 3: Write f'(x) as sum of three determinants</strong></p><p>f'(x) = |0 2 6x | + |1 2x 3x² | + |1 2x 3x² |</p><p> |x x² x³ | |1 2x 3x² | |x x² x³ |</p><p> |0 2 6x | |0 2 6x | |0 0 6 |</p><p><strong>Step 4: Evaluate at x = 1</strong></p><p>f'(1) = |0 2 6| + |1 2 3| + |1 2 3|</p><p> |1 1 1| |1 2 3| |1 1 1|</p><p> |0 2 6| |0 2 6| |0 0 6|</p><p><strong>Step 5: Calculate the first determinant (D₁)</strong></p><p>D₁ = |0 2 6|</p><p> |1 1 1| = 0 - 2(6-0) + 6(2-0) = 0 - 12 + 12 = 0</p><p> |0 2 6|</p><p><strong>Step 6: Calculate the second determinant (D₂)</strong></p><p>D₂ = |1 2 3|</p><p> |1 2 3| = 0 (rows 1 and 2 are identical)</p><p> |0 2 6|</p><p><strong>Step 7: Calculate the third determinant (D₃)</strong></p><p>D₃ = |1 2 3|</p><p> |1 1 1| = 1(6-0) - 2(6-0) + 3(0-0) = 6 - 12 = -6</p><p> |0 0 6|</p><p><strong>Step 8: Sum the determinants</strong></p><p>f'(1) = D₁ + D₂ + D₃ = 0 + 0 + (-6) = -6</p><p><strong>∴ Answer: -6</strong></p>
Correct Answer: -6