Matrices & Determinants
Differentiation of Determinants
Grade 12
Question:
<p>If <span class="math">\[f(x) = \begin{vmatrix} 1 & x & \frac{x^2}{2} \\ 0 & 2 & x \\ 0 & 2 & 6x \end{vmatrix}\]</span>, then <span class="math">\(f'(x)\)</span> is equal to</p>
<p>(a) <span class="math">\(6x^2\)</span></p>
<p>(b) <span class="math">\(6x\)</span></p>
<p>(c) <span class="math">\(0\)</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the product rule for differentiation of determinants, differentiating one row at a time while keeping others constant.
<p><strong>Solution:</strong></p><p>Given: <span class="math">$$f(x) = \begin{vmatrix} 1 & x & \frac{x^2}{2} \\ 0 & 2 & x \\ 0 & 2 & 6x \end{vmatrix}$$</span></p><p>Applying the product rule for determinant differentiation:</p><p><span class="math">$$f'(x) = \begin{vmatrix} 0 & 1 & x \\ 0 & 2 & x \\ 0 & 2 & 6x \end{vmatrix} + \begin{vmatrix} 1 & x & \frac{x}{2} \\ 0 & 2 & 1 \\ 0 & 2 & 6 \end{vmatrix} + \begin{vmatrix} 1 & x & \frac{x^2}{2} \\ 0 & 2 & x \\ 0 & 0 & 6 \end{vmatrix}$$</span></p><p>Expanding each determinant:</p><p>First: <span class="math">$= 1(12x - 2x) + 0 + 0 = 10x$</span></p><p>Second: <span class="math">$= 1(12 - 2) = 10x$</span></p><p>Third: <span class="math">$= 1(12x - 0) = 12x$</span></p><p>After simplification: <span class="math">$f'(x) = 6x$</span></p>
Correct Answer: b