We find the value of the constant by applying the column-by-column or row-by-row rule for differentiating determinants and substituting the variable point.
1. Recall the differentiation rule for a $$3 \times 3$$ determinant layout. The derivative $$D'(x)$$ is the sum of three independent determinants, where we differentiate one row (or column) at a time while leaving the other rows (or columns) unchanged:
$$D'(x) = \begin{vmatrix} \frac{d}{dx}(1+x^2) & \frac{d}{dx}(x) & \frac{d}{dx}(x^2) \\ x & 1+x & x \\ x^2 & x & 1+x^2 \end{vmatrix} + \begin{vmatrix} 1+x^2 & x & x^2 \\ \frac{d}{dx}(x) & \frac{d}{dx}(1+x) & \frac{d}{dx}(x) \\ x^2 & x & 1+x^2 \end{vmatrix} + \begin{vmatrix} 1+x^2 & x & x^2 \\ x & 1+x & x \\ \frac{d}{dx}(x^2) & \frac{d}{dx}(x) & \frac{d}{dx}(1+x^2) \end{vmatrix}$$
2. Compute the individual derivative terms inside each row slot:
$$D'(x) = \begin{vmatrix} 2x & 1 & 2x \\ x & 1+x & x \\ x^2 & x & 1+x^2 \end{vmatrix} + \begin{vmatrix} 1+x^2 & x & x^2 \\ 1 & 1 & 1 \\ x^2 & x & 1+x^2 \end{vmatrix} + \begin{vmatrix} 1+x^2 & x & x^2 \\ x & 1+x & x \\ 2x & 1 & 2x \end{vmatrix}$$
3. Substitute $$x = 0$$ directly into each of the three component determinants to evaluate $$D'(0)$$:
$$D'(0) = \begin{vmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{vmatrix}$$
4. Evaluate each numerical matrix determinant explicitly:
* First matrix determinant: The first column consists entirely of zeros, so its determinant evaluates to $$0$$.
* Second matrix determinant: Expand along the first row:
$$1 \cdot \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} - 0 + 0 = 1 \cdot (1 - 0) = 1$$
* Third matrix determinant: The third row is identical to the second row layout after substitution ($$[0, 1, 0]$$), which forces its determinant to evaluate to $$0$$.
5. Sum the values of the three parts together:
$$D'(0) = 0 + 1 + 0 = 1$$
Thus, the value of the constant $$k$$ is exactly $$1$$.
91- 6303239042
