Matrices & Determinants
Differentiation of Determinants
Grade 12
Question:
<p>Given \(f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{vmatrix}\), find \(f'(x)\).</p>
<p>\(\begin{vmatrix} -\sin x & 1 & 0 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{vmatrix} + \begin{vmatrix} \cos x & x & 1 \\ 2\cos x & 2x & 2 \\ \tan x & x & 1 \end{vmatrix} + \begin{vmatrix} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \sec^2 x & 1 & 0 \end{vmatrix}\)</p>
<p>Sum of determinants with differentiated rows</p>
<p>0</p>
<p>\(2\cos x - \sin x\)</p>
Step-by-Step Solution
Key Concept: Use the linearity property of determinant differentiation: the derivative of a determinant is the sum of determinants where each row is differentiated in turn, avoiding the product rule on the entire determinant.
<p><strong>Step 1:</strong> Apply the determinant differentiation rule: f'(x) = det(A₁) + det(A₂) + det(A₃), where Aᵢ has the i-th row differentiated.</p><p><strong>Step 2:</strong> Differentiate Row 1: (-sin x, 1, 0)</p><p>det(A₁) = ∣-sin x 1 0∣ = -sin x(x² - 2x) - 1(2sin x - 2x tan x)</p><p> ∣2sin x x² 2x∣</p><p> ∣tan x x 1∣</p><p><strong>Step 3:</strong> Differentiate Row 2: (2cos x, 2x, 2)</p><p>det(A₂) = ∣cos x x 1∣ = cos x(2x - 2x) - x(2cos x - 2tan x) + 1(2sin x cos x - 2x tan x)</p><p> ∣2cos x 2x 2∣</p><p> ∣tan x x 1∣</p><p><strong>Step 4:</strong> Differentiate Row 3: (sec² x, 1, 0)</p><p>det(A₃) = ∣cos x x 1∣ = cos x(x² - 2x) - x(2sin x - 2x tan x) + 1(2sin x - x² sec² x)</p><p> ∣2sin x x² 2x∣</p><p> ∣sec² x 1 0∣</p><p><strong>Step 5:</strong> Observe that f(x) has identical Row 1 and Row 3 (cos x, x, 1), so f(x) = 0 identically.</p><p>∴ f'(x) = 0</p><p><strong>Answer: A</strong></p>
Correct Answer: A