Am I doing it wrong or this derivative is really lengthy? We have:
$$f(x)=\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\sin x& x^2 & 2x \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}$$
And we are supposed to calculate $$\lim_{x\rightarrow 0}\frac{1}{x^6}\int_{x^2}^{x^3}f(x).dx=A$$
and tell if $[A]$ (where $[x]$ is greatest integer less than or equal to $x$) is equal to $f''(0)$ or not.
Yes, i know how to differentiate a determinant But To be honest , this question looks barabaric in cases where I need to solve it under $5$ minutes or $8$ max.Does there exist any simpler way than 
$$f'(x)=\begin{vmatrix} \alpha \cos^2 x+(\alpha x+1)(-2\sin x \cos x) & 1 & -1\\\beta\sin x& x^2 & 2x \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}+\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\cos  x& 2x & 2 \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}+\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\sin x& x^2 & 2x \\2\gamma x \tan x+ \sec^2 (\gamma x^2+1) & 1 & -2x\end{vmatrix}$$
and proceeding?
Please guide.

Let $(u,v,w)$ be the entries in $1^{st}$ column of the determinant $f(x)$, we have
$$f(x) 
= \begin{vmatrix} u & x & 1-x\\ v & x^2 & 2x \\ w & x & 1-x^2\end{vmatrix}
= x\begin{vmatrix} u & 1 & 1-x\\ v & x & 2x \\ w & 1 & 1-x^2\end{vmatrix}
= x\begin{vmatrix} u & 1 & -x\\ v & x & x \\ w & 1 & -x^2\end{vmatrix}
= x^2\begin{vmatrix} u & 1 & -1\\ v & x & 1 \\ w & 1 & -x\end{vmatrix}
$$
Let $g(x)$ be the rightmost determinant. Expand it as a power series in $x$,
$$g(x) = \begin{vmatrix} u & 1 & -1\\ v & x & 1 \\ w & 1 & -x\end{vmatrix} = g_0  + g_1 x + g_2 x^2 + \cdots$$
and substitute it in the integral, we get
$$\begin{align}
\frac{1}{x^6}\int_{x^2}^{x^3} f(t) dt
&= \frac{1}{x^6}\int_{x^2}^{x^3} t^2 g(t) dt
= \frac{1}{x^6}\int_{x^2}^{x^3} ( g_0 t^2 + g_1 t^3 + \cdots ) dt\\
&= \frac{1}{x^6}\left( \frac{g_0}{3} [t^3]_{x^2}^{x^3} + \frac{g_1}{4} [t^4]_{x^2}^{x^3} + \cdots\right)
= \frac{1}{x^6}\left( \frac{g_0}{3} (x^9 - x^6) + \frac{g_1}{4} (x^{12}-x^8) + \cdots\right)\\
&= -\frac13 g_0 + O(x^2)
\end{align}
$$
Notice when $x = 0$, $(u,v,w) = (1,0,0)$, we find
$$A = \lim_{x\to0} \left(-\frac{g_0}{3} + O(x^2)\right) 
= -\frac{g_0}{3} =
-\frac13 
\begin{vmatrix} 1 & 1 & -1\\ 0 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix}
= \frac13
$$
This leads to
$$\left\lfloor A \right\rfloor = 0\quad\ne\quad f''(0) = 2g(0) = 2g_0 = -2$$
