If `Delta=|(x,x^2,x^3),(1,2x,3x^2),(0,2,6x)|` then `d/(dx)(Delta)=`

Options

A. 6
B. 6x
C. `6x^2`
D. 0

Answer

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

\[ \Delta = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix} \]

### Step 1: Factor out common terms from rows
We can simplify the determinant by factoring out common terms from the rows.

- From the first row, we can take out \(x\).
- From the second row, we can take out \(2\).
- From the third row, we can take out \(2\).

Thus, we have:

\[ \Delta = x \cdot 2 \cdot 2 \cdot \begin{vmatrix} 1 & x & x^2 \\ 1 & 2x & 3x^2 \\ 0 & 1 & 3 \end{vmatrix} \]

This simplifies to:

\[ \Delta = 4x \cdot \begin{vmatrix} 1 & x & x^2 \\ 1 & 2x & 3x^2 \\ 0 & 1 & 3 \end{vmatrix} \]

### Step 2: Calculate the determinant
Now we need to calculate the determinant:

\[ \begin{vmatrix} 1 & x & x^2 \\ 1 & 2x & 3x^2 \\ 0 & 1 & 3 \end{vmatrix} \]

Using the first column for expansion, we have:

\[ = 1 \cdot \begin{vmatrix} 2x & 3x^2 \\ 1 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} x & x^2 \\ 1 & 3 \end{vmatrix} + 0 \]

Calculating the first determinant:

\[ \begin{vmatrix} 2x & 3x^2 \\ 1 & 3 \end{vmatrix} = (2x)(3) - (3x^2)(1) = 6x - 3x^2 = 3x(2 - x) \]

Calculating the second determinant:

\[ \begin{vmatrix} x & x^2 \\ 1 & 3 \end{vmatrix} = (x)(3) - (x^2)(1) = 3x - x^2 = x(3 - x) \]

Putting it all together:

\[ \Delta = 4x \left( 3x(2 - x) - x(3 - x) \right) \]

### Step 3: Simplify the expression
Now simplify the expression inside the parentheses:

\[ = 4x \left( 6x - 3x^2 - 3x + x^2 \right) = 4x \left( 3x - 2x^2 \right) \]

Thus, we have:

\[ \Delta = 12x^2 - 8x^3 \]

### Step 4: Differentiate with respect to \(x\)
Now we differentiate \(\Delta\) with respect to \(x\):

\[ \frac{d}{dx}(\Delta) = \frac{d}{dx}(12x^2 - 8x^3) \]

Using the power rule:

\[ = 24x - 24x^2 \]

### Final Answer
Thus, the final answer is:

\[ \frac{d}{dx}(\Delta) = 24x - 24x^2 \]