If y = `|{:(sin x , cos x , sin x),(cos x,-sin x, cos x),(x, 1, 1):}|`, then `(dy)/(dx)` is

Options

A. 0
B. 1
C. `-1`
D. None ot these

Answer

To solve the problem, we need to find the derivative of the determinant given by:

\[ y = \begin{vmatrix} \sin x & \cos x & \sin x \\ \cos x & -\sin x & \cos x \\ x & 1 & 1 \end{vmatrix} \]

### Step 1: Expand the Determinant

We will expand the determinant using the formula for 3x3 determinants:

\[ y = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

For our determinant:

\[ y = \sin x \begin{vmatrix} -\sin x & \cos x \\ 1 & 1 \end{vmatrix} - \cos x \begin{vmatrix} \cos x & \cos x \\ x & 1 \end{vmatrix} + \sin x \begin{vmatrix} \cos x & -\sin x \\ x & 1 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

1. For the first determinant:

\[ \begin{vmatrix} -\sin x & \cos x \\ 1 & 1 \end{vmatrix} = (-\sin x)(1) - (\cos x)(1) = -\sin x - \cos x \]

2. For the second determinant:

\[ \begin{vmatrix} \cos x & \cos x \\ x & 1 \end{vmatrix} = (\cos x)(1) - (\cos x)(x) = \cos x - x \cos x = \cos x(1 - x) \]

3. For the third determinant:

\[ \begin{vmatrix} \cos x & -\sin x \\ x & 1 \end{vmatrix} = (\cos x)(1) - (-\sin x)(x) = \cos x + x \sin x \]

### Step 3: Substitute Back into the Determinant Expression

Now substituting back into the expression for \( y \):

\[ y = \sin x (-\sin x - \cos x) - \cos x (\cos x(1 - x)) + \sin x (\cos x + x \sin x) \]

### Step 4: Simplify the Expression

Now we simplify:

\[ y = -\sin^2 x - \sin x \cos x - \cos^2 x(1 - x) + \sin x \cos x + x \sin^2 x \]

Combining like terms:

\[ y = -\sin^2 x - \cos^2 x + x \sin^2 x \]

Using the identity \( \sin^2 x + \cos^2 x = 1 \):

\[ y = x \sin^2 x - 1 \]

### Step 5: Differentiate \( y \)

Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}(x \sin^2 x - 1) \]

Using the product rule:

\[ \frac{dy}{dx} = \sin^2 x + x \cdot 2 \sin x \cos x \]

Thus, we have:

\[ \frac{dy}{dx} = \sin^2 x + 2x \sin x \cos x \]

### Final Answer

The derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \sin^2 x + 2x \sin x \cos x \]