Question:
If \[ f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}, \] then \[ f'(x) \] is equal to:
If \[ f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}, \] then \[ f'(x) \] is equal to:
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Always simplify the determinant first before differentiating — it saves time and avoids mistakes.
Updated On: May 1, 2026
- \( x^3 + 6x^2 \)
- \( 6x^3 \)
- \( 3x \)
- \( 6x^2 \)
- \( 0 \)
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The Correct Option is B
Solution and Explanation
Concept: First evaluate the determinant to get \( f(x) \) as a polynomial, then differentiate normally.
Step 1: Expand the determinant.
\[ f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix} \] Expand along first row: \[ = x \begin{vmatrix} 2x & 3x^2 \\ 2 & 6x \end{vmatrix} - x^2 \begin{vmatrix} 1 & 3x^2 \\ 0 & 6x \end{vmatrix} + x^3 \begin{vmatrix} 1 & 2x \\ 0 & 2 \end{vmatrix} \]
Step 2: Evaluate minors.
First minor: \[ (2x)(6x) - (3x^2)(2) = 12x^2 - 6x^2 = 6x^2 \] Second minor: \[ (1)(6x) - (3x^2)(0) = 6x \] Third minor: \[ (1)(2) - (2x)(0) = 2 \]
Step 3: Substitute back.
\[ f(x) = x(6x^2) - x^2(6x) + x^3(2) \] \[ = 6x^3 - 6x^3 + 2x^3 \] \[ = 2x^3 \]
Step 4: Differentiate.
\[ f'(x) = \frac{d}{dx}(2x^3) = 6x^2 \]
Step 5: Final answer.
\[ \boxed{6x^2} \]
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