Derivatives using matrices good $$\left|\begin{matrix}
(1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\
(1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\
(1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\
\end{matrix}\right |$$
Find the coefficient of $x$ in the above determinant.
Try using differentiability.

Denote the determinant by $f(x)$. Then the $x$ coefficient is given by $f'(0)$. Now the derivative of a determinant equals the sum of determinants of the matrices obtained by taking the derivative of each row separately. Hence
$$
\begin{align}
f'(0)
&=
\frac{\mathrm{d}}{\mathrm{d}x}
\left|
  \begin{matrix}
    (1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\
    (1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\
    (1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\
  \end{matrix}
\right|_{x=0} \\
&=
\left|
  \begin{matrix}
    a_1b_1 & a_1b_2 & a_1b_3 \\
    1      & 1      & 1 \\
    1      & 1      & 1 \\
  \end{matrix}
\right|
+
\left|
  \begin{matrix}
    1      & 1      & 1 \\
    a_2b_1 & a_2b_2 & a_2b_3 \\
    1      & 1      & 1 \\
  \end{matrix}
\right|
+
\left|
  \begin{matrix}
    1      & 1      & 1 \\
    1      & 1      & 1 \\
    a_3b_1 & a_3b_2 & a_3b_3 \\
  \end{matrix}
\right| \\
&= 0
\end{align}
$$
