Question:
Let \( f_1(x), f_2(x), g_1(x), g_2(x) \) be differentiable functions on \( \mathbb{R} \). Let
\[
F(x) = \begin{vmatrix} f_1(x) & f_2(x) \\
g_1(x) & g_2(x) \end{vmatrix}.
\]
Then \( F'(x) \) is equal to
Let \( f_1(x), f_2(x), g_1(x), g_2(x) \) be differentiable functions on \( \mathbb{R} \). Let
\[ F(x) = \begin{vmatrix} f_1(x) & f_2(x) \\ g_1(x) & g_2(x) \end{vmatrix}. \] Then \( F'(x) \) is equal to
\[ F(x) = \begin{vmatrix} f_1(x) & f_2(x) \\ g_1(x) & g_2(x) \end{vmatrix}. \] Then \( F'(x) \) is equal to
Show Hint
When differentiating a 2x2 matrix determinant, apply the rule for matrix determinants and differentiate each element with respect to \( x \).
Updated On: Nov 20, 2025
- \( f_1'(x) f_2'(x) + f_1(x) g_1'(x) \)
- \( f_1'(x) f_2'(x) - |f_1(x) g_1'(x)| \)
- \( f_1'(x) f_2'(x) + f_1(x) g_1'(x) \)
- \( f_1'(x) g_2'(x) - f_2'(x) g_2'(x) \)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the determinant.
We are given a determinant of a 2×2 matrix:
F(x) = | f₁(x) f₂(x) |
| g₁(x) g₂(x) |
The derivative of this determinant with respect to x is obtained by differentiating:
F(x) = f₁(x)g₂(x) − f₂(x)g₁(x)
Step 2: Apply the derivative rule.
Differentiating each term gives:
F′(x) = f₁′(x)g₂(x) + f₁(x)g₂′(x) − f₂′(x)g₁(x) − f₂(x)g₁′(x)
Step 3: Conclusion.
Thus, the correct answer is (C).
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