How to calculate this pretty determinant smartly? Let $$f(x):=a_1+a_2 \sin(x)+a_3 \sin(x)^2$$
$$g(x):=b_1+b_2 \sin(x)+b_3 \sin(x)^2$$
$$h(x):=c_1+c_2 \sin(x)+c_3 \sin(x)^2$$
then Mathematica shows that the determinant of the matrix
$$\left(\begin{matrix} f(x) & g(x)& h(x) \\ f'(x) & g'(x) & h'(x) \\ f''(x) & g''(x) & h''(x) \end{matrix} \right)$$ is just
$$2 (-a_3 b_2 c_1 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_1 b_3 c_2 - a_2 b_1 c_3 + 
   a_1 b_2 c_3) \cos(x)^2$$
which is a suprisingly simple result given that the expressions from the chain-rule may become rather cumbersome. 
I would like to know, is there a smart way to conclude this result without explicitly calculating everything and regrouping terms in order to see the result?
 A: This becomes simpler if we let
$$u(x) = 1 \qquad v(x) = \sin(x) \qquad w(x) = \sin^2(x)$$
so since differentiation is linear, 
$$\left(\begin{matrix} f(x) \\ f'(x) \\ f''(x) \end{matrix} \right)
= a_1 \left(\begin{matrix} u(x) \\ u'(x) \\ u''(x) \end{matrix} \right)
+ a_2 \left(\begin{matrix} v(x) \\ v'(x) \\ v''(x) \end{matrix} \right)
+ a_3 \left(\begin{matrix} w(x) \\ w'(x) \\ w''(x) \end{matrix} \right)$$
and similar formulas hold for $g$ in terms of $b_i$'s and $h$ in terms of $c_i$'s. From here we can write
$$\left(\begin{matrix} f(x) & g(x)& h(x) \\ f'(x) & g'(x) & h'(x) \\ f''(x) & g''(x) & h''(x) \end{matrix} \right)
= 
\left(\begin{matrix} u(x) & v(x)& w(x) \\ u'(x) & v'(x) & w'(x) \\ u''(x) & v''(x) & w''(x) \end{matrix} \right)
\left(\begin{matrix} a_1 & b_1& c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{matrix} \right).
$$
Call these matrices $F$, $U$, $A$, so this equation reads $F = UA$. We conclude $\det F = \det U \det A$. 
We can easily compute
$$ \det U = \det \left(\begin{matrix} 1 & \sin(x) & \sin^2(x) \\ 0 & \cos(x) & 2\cos(x)\sin(x) \\ 0 & -\sin(x) & 2(\cos^2(x) - \sin^2(x)) \end{matrix} \right)
 = 2\cos^3(x)$$
so $\det F = 2 \cos^3(x) \det A$.
A: You have $$f(x)=p(\sin x),\ \ g(x)=q(\sin x),\ \ h(x)=r(\sin x)$$ for three degree-two polynomials. Then
$$
f'(x)=p'(\sin x)\cos x,\ \ f''(x)=p''(\sin x)\cos x-p'(x)\sin x.
$$
So the determinant is 
\begin{align}
\begin{vmatrix}
p(\sin x)&q(\sin x)& r(\sin  x)\\ 
p'(\sin x)\cos x&q'(\sin x)\cos x& r'(\sin x)\cos x\\
p''(\sin x)\cos^2 x-p'(\sin x)\sin x&q''(\sin x)\cos^2 x-q'(\sin x)\sin x&r''(\sin x)\cos^2 x-r'(\sin x)\sin x
\end{vmatrix}\\
\end{align}
We can decompose this as the sum of
$$
\begin{vmatrix}
p(\sin x)&q(\sin x)& r(\sin  x)\\ 
p'(\sin x)\cos x&q'(\sin x)\cos x& r'(\sin x)\cos x\\
p''(\sin x)\cos^2 x &q''(\sin x)\cos^2 x &r''(\sin x)\cos^2 x 
\end{vmatrix}
=
c^3\,\begin{vmatrix}
p(\sin x)&q(\sin x)& r(\sin  x)\\ 
p'(\sin x) &q'(\sin x) & r'(\sin x) \\
p''(\sin x)  &q''(\sin x) &r''(\sin x)  
\end{vmatrix},
$$
where $c=\cos x$,
and
\begin{align}
\begin{vmatrix}
p(\sin x)&q(\sin x)& r(\sin  x)\\ 
p'(\sin x)\cos x&q'(\sin x)\cos x& r'(\sin x)\cos x\\
-p'(\sin x)\sin x& -q'(\sin x)\sin x& -r'(\sin x)\sin x
\end{vmatrix}=0
\end{align}
For the determinant with the polynomials, we have 
$$
\begin{vmatrix}
a_1+a_2s+a_3s^2& b_1+b_2s+b_3s^2& c_1+c_2s+c_3s^2\\
a_2+2a_3s&b_2+2b_3s&c_2+2c_3s\\
2a_3&2b_3&2b_3
\end{vmatrix} 
$$
Doing row reduction (first row minus $s$ times the second, and second row minus $s$ times the third), this last determinant equals 
$$
\begin{vmatrix}
a_1& b_1& c_1\\
a_2&b_2&c_2\\
2a_3&2b_3&2b_3
\end{vmatrix} 
$$
So the determinant is equal to 
$$
\begin{vmatrix}
a_1& b_1& c_1\\
a_2&b_2&c_2\\
2a_3&2b_3&2b_3
\end{vmatrix} \,\cos^3x
$$
A: The critical step to doing this 'smartly' is to realize that you can write your determinant as the product of two matrices as follows. We also know that $\det(A\cdot B)=\det(A)\cdot \det(B)$, so we can write the determinant as
$$
\begin{vmatrix}
1&\sin{x}&\sin^2{x}\\
0&\cos{x}&\sin{2x}\\
0&-\sin{x}&2\cos{2x}
\end{vmatrix}
\cdot
\begin{vmatrix}
a_1& b_1& c_1\\
a_2&b_2&c_2\\
a_3&b_3&b_3
\end{vmatrix}
$$
Now we just need to find the determinant of the matrix on the left, which is much simpler than what we were doing before, and turns out just to be $2\cos^2{x}$
