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**Hint:**Differentiation: Differentiation is the area of change with respect to the input.

The value of differentiation of a constant term is always zero.

Product rule of differentiation: Let us consider \[f(x),g(x)\] be the function of \[x.\]

Then, \[\dfrac{d}{{dx}}[f(x)g(x)] = \dfrac{d}{{dx}}[f(x)]g(x) + f(x)\dfrac{d}{{dx}}[g(x)] = f'(x)g(x) + f(x)g'(x)\]

**Complete step-by-step answer:**

It is given that,

\[F = \det \left( {\begin{array}{*{20}{c}}

{{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\

{{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\

{{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)}

\end{array}} \right)\]

Where, \[{f_r}(x),{g_r}(x),{h_r}(x),r = 1,2,3\] are polynomials in\[x\].

Differentiate \[F\] with respect to \[x\] we get,

\[F' = \det \left( {\begin{array}{*{20}{c}}

{f{'_1}(x)}&{f{'_2}(x)}&{f{'_3}(x)} \\

{{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\

{{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)}

\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}

{{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\

{{g_1}'(x)}&{{g_2}'(x)}&{{g_3}'(x)} \\

{{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)}

\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}

{{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\

{{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\

{h{'_1}(x)}&{h{'_2}(x)}&{h{'_3}(x)}

\end{array}} \right)\]\[x = a\]

Substitute in \[F\] we get,

\[F' = \det \left( {\begin{array}{*{20}{c}}

{f{'_1}(a)}&{f{'_2}(a)}&{f{'_3}(a)} \\

{{g_1}(a)}&{{g_2}(a)}&{{g_3}(a)} \\

{{h_1}(a)}&{{h_2}(a)}&{{h_3}(a)}

\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}

{f{'_1}(a)}&{f{'_2}(a)}&{f{'_3}(a)} \\

{{g_1}(a)}&{{g_2}(a)}&{{g_3}(a)} \\

{{h_1}(a)}&{{h_2}(a)}&{{h_3}(a)}

\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}

{{f_1}(a)}&{{f_2}(a)}&{{f_3}(a)} \\

{{g_1}(a)}&{{g_2}(a)}&{{g_3}(a)} \\

{h{'_1}(a)}&{h{'_2}(a)}&{h{'_3}(a)}

\end{array}} \right)\]

As per the given condition, \[{f_r}(a) = {g_r}(a) = {h_r}(a),r = 1,2,3\]

Also we know by following the property of determinant, \[\det \left( {\begin{array}{*{20}{c}}

a&b&c \\

a&b&c \\

a&b&c

\end{array}} \right) = 0\]we get,

\[F' = 0\]

**Hence, \[F'(x)\] at \[x = a\] is \[0.\]**

**Note:**The determinant of a matrix is a special number that can be calculated from a square matrix. The determinant helps us to find the inverse of a matrix.

Differentiation helps us to find rates of change. For example, it helps us to find the rate of change of velocity with respect to time (which is known as acceleration). It also allows us to find the rate of change of x with respect to y, which on a graph of y against x is the gradient of the curve (which is known as slope). There are a number of simple rules which can be used to allow us to differentiate many functions easily.

If y = some function of x (in other words if y is equal to an expression containing numbers and x's), then the derivative of y (with respect to x) is pronounced "dee y by dee x”.

The differentiation of matrices is the place where every one of us would make mistakes so we should be very careful while doing it.

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