Question:
If
\[
y=
\begin{vmatrix}
\sin x & \cos x & \sin x\\
\cos x & -\sin x & \cos x\\
x & 1 & 1
\end{vmatrix},
\]
then
\[
\frac{dy}{dx}=
\]
If
\[
y=
\begin{vmatrix}
\sin x & \cos x & \sin x\\
\cos x & -\sin x & \cos x\\
x & 1 & 1
\end{vmatrix},
\]
then
\[
\frac{dy}{dx}=
\]
Show Hint
Before differentiating a determinant expression, always try to simplify it using row or column operations.
Updated On: Jun 3, 2026
- $x\cos x$
- $x\sin x$
- $\sin x+\cos x$
- $1$
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Simplify the determinant before differentiating.
Step 2: Meaning
Perform \[ C_3\to C_3-C_1. \] Then the third column becomes \[ \begin{bmatrix} 0\\0\\1-x \end{bmatrix}. \]
Step 3: Analysis
Expanding along the third column, \[ y=(1-x) \begin{vmatrix} \sin x & \cos x\\ \cos x & -\sin x \end{vmatrix}. \] Now \[ \begin{vmatrix} \sin x & \cos x\\ \cos x & -\sin x \end{vmatrix} = -\sin^2x-\cos^2x=-1. \] Hence \[ y=(1-x)(-1)=x-1. \]
Step 4: Conclusion
Therefore \[ \frac{dy}{dx}=1. \]
Final Answer: (D)
Simplify the determinant before differentiating.
Step 2: Meaning
Perform \[ C_3\to C_3-C_1. \] Then the third column becomes \[ \begin{bmatrix} 0\\0\\1-x \end{bmatrix}. \]
Step 3: Analysis
Expanding along the third column, \[ y=(1-x) \begin{vmatrix} \sin x & \cos x\\ \cos x & -\sin x \end{vmatrix}. \] Now \[ \begin{vmatrix} \sin x & \cos x\\ \cos x & -\sin x \end{vmatrix} = -\sin^2x-\cos^2x=-1. \] Hence \[ y=(1-x)(-1)=x-1. \]
Step 4: Conclusion
Therefore \[ \frac{dy}{dx}=1. \]
Final Answer: (D)
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