Question:

Let the given function be $$f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x\end{vmatrix}$$ Observe that the second row is the derivative of the first row, and the third row is the derivative of the second row. To find $f'(1)$, first expand the determinant to express $f(x)$ as a polynomial in $x$. Expanding along the first row: $$f(x) = x \begin{vmatrix} 2x & 3x^2 \\ 2 & 6x \end{vmatrix} - x^2 \begin{vmatrix} 1 & 3x^2 \\ 0 & 6x \end{vmatrix} + x^3 \begin{vmatrix} 1 & 2x \\ 0 & 2 \end{vmatrix}$$ $$f(x) = x((2x)(6x) - (3x^2)(2)) - x^2((1)(6x) - (3x^2)(0)) + x^3((1)(2) - (2x)(0))$$ $$f(x) = x(12x^2 - 6x^2) - x^2(6x - 0) + x^3(2 - 0)$$ $$f(x) = x(6x^2) - x^2(6x) + x^3(2)$$ $$f(x) = 6x^3 - 6x^3 + 2x^3$$ $$f(x) = 2x^3$$ Now, differentiate $f(x)$ with respect to $x$: $$f'(x) = \frac{d}{dx}(2x^3)$$ $$f'(x) = 6x^2$$ Finally, evaluate $f'(x)$ at $x=1$: $$f'(1) = 6(1)^2$$ $$f'(1) = 6$$