<p>Let <br>\[f(x) = \begin{vmatrix} \cos(x+x^2) & \sin(x+x^2) & -\cos(x+x^2) \\ \sin(x-x^2) & \cos(x-x^2) & \sin(x-x^2) \\ \sin 2x & 0 & \sin(2x^2) \end{vmatrix}\]<br>Find the value of \(f'(0)\).</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is a determinant function; use the product rule for derivatives of determinants: f'(x) = sum of determinants where each row is differentiated one at a time. At x=0, evaluate the structure to identify which differentiated terms survive.
<p><strong>Step 1:</strong> Use the derivative rule for determinants. If f(x) is a determinant with rows R₁, R₂, R₃, then:</p><p>f'(x) = det(R₁', R₂, R₃) + det(R₁, R₂', R₃) + det(R₁, R₂, R₃')</p><p><strong>Step 2:</strong> Evaluate each row at x=0:</p><p>R₁(0) = [cos(0), sin(0), -cos(0)] = [1, 0, -1]</p><p>R₂(0) = [sin(0), cos(0), sin(0)] = [0, 1, 0]</p><p>R₃(0) = [sin(0), 0, sin(0)] = [0, 0, 0]</p><p><strong>Step 3:</strong> Compute derivatives at x=0:</p><p>R₁'(x) = [-(1+2x)sin(x+x²), (1+2x)cos(x+x²), (1+2x)sin(x+x²)]</p><p>R₁'(0) = [0, 1, 0]</p><p>R₂'(x) = [(1-2x)cos(x-x²), -(1-2x)sin(x-x²), (1-2x)cos(x-x²)]</p><p>R₂'(0) = [1, 0, 1]</p><p>R₃'(x) = [2cos(2x), 0, 4x·cos(2x²)]</p><p>R₃'(0) = [2, 0, 0]</p><p><strong>Step 4:</strong> Calculate the three determinants at x=0:</p><p>det(R₁', R₂, R₃) = det([0,1,0; 0,1,0; 0,0,0]) = 0</p><p>det(R₁, R₂', R₃) = det([1,0,-1; 1,0,1; 0,0,0]) = 0</p><p>det(R₁, R₂, R₃') = det([1,0,-1; 0,1,0; 2,0,0]) = 1(1·0 - 0·0) - 0 + (-1)(0·0 - 1·2) = 0 + 2 = 2</p><p><strong>Step 5:</strong> f'(0) = 0 + 0 + 2 = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2