If we have a determinant as $y = \left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right|$ then
Prove that $\dfrac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}
  {f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right|$

Hint: In this particular question use the concept that the differentiation of determinant is the sum of three determinants, in first determinant, differentiation of first row w.r.t x and second and third row remain same, in second determinant, differentiation of second row w.r.t x and first and third row remain same, in third determinant, differentiation of third row w.r.t x and first and second row remain same, so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given determinant
$y = \left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right|$.............. (1)
Now we have to prove that $\dfrac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}
  {f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right|$
Proof –
Differentiate equation (1) w.r.t x both sides we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right|$
So according to the property of determinant if a differential operator is outside the determinant and we want this operator to go inside the determinant then this will happen in a particular way.
It is the sum of three determinants, in first determinant, differentiation of first row w.r.t x and second and third row remain same, in second determinant, differentiation of second row w.r.t x and first and third row remain same, in third determinant, differentiation of third row w.r.t x and first and second row remain same, so we have
$ \Rightarrow \dfrac{d}{{dx}}y = \left| {\begin{array}{*{20}{c}}
  {\dfrac{d}{{dx}}f\left( x \right)}&{\dfrac{d}{{dx}}g\left( x \right)}&{\dfrac{d}{{dx}}h\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  {\dfrac{d}{{dx}}l}&{\dfrac{d}{{dx}}m}&{\dfrac{d}{{dx}}n} \\
  a&b&c
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  l&m&n \\
  {\dfrac{d}{{dx}}a}&{\dfrac{d}{{dx}}b}&{\dfrac{d}{{dx}}c}
\end{array}} \right|$
Now as we know that l, m, n, a, b, and c are constant w.r.t x so its differentiation is zero w.r.t x so we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left| {\begin{array}{*{20}{c}}
  {f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  0&0&0 \\
  a&b&c
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)} \\
  l&m&n \\
  0&0&0
\end{array}} \right|$
Where $\left[ {\dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right),\dfrac{d}{{dx}}g\left( x \right) = g'\left( x \right),\dfrac{d}{{dx}}h\left( x \right) = h'\left( x \right)} \right]$
Now as we know that in a determinant if one complete row is zero than the value of whole determinant is zero so we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left| {\begin{array}{*{20}{c}}
  {f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right| + 0 + 0$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}
  {f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)} \\
  l&m&n \\
  a&b&c
\end{array}} \right|$
Hence proved.

Note: Whenever we face such types of questions the key concept we have to remember is the property how to differentiate the determinate which is all stated above without this property we cannot solve these problems, this is the basic building step of the problem so differentiate it as above and simplify we will get the required answer.