If $y = \left| {\begin{array}{llllllllllllllllllll}{f(x)}&{g(x)}&{h(x)}\\l&m&n\\a&b&c\end{array}} \right|,$ prove that
$\cfrac{{dy}}{{dx}} = \left| {\begin{array}{llllllllllllllllllll}{f'(x)}&{g'(x)}&{h'(x)}\\l&m&n\\a&b&c\end{array}} \right|$ .
If $y = \left| {\begin{array}{llllllllllllllllllll}{f(x)}&{g(x)}&{h(x)}\\l&m&n\\a&b&c\end{array}} \right|,$ prove that
$\cfrac{{dy}}{{dx}} = \left| {\begin{array}{llllllllllllllllllll}{f'(x)}&{g'(x)}&{h'(x)}\\l&m&n\\a&b&c\end{array}} \right|$ .
Official Solution
We have , $y = \left| {\begin{array}{llllllllllllllllllll}{f(x)}&{g(x)}&{h(x)}\\l&m&n\\a&b&c\end{array}} \right|$
$\therefore \cfrac{{dy}}{{dx}} = \left| {\begin{array}{llllllllllllllllllll}{\cfrac{d}{{dx}}(f(x))}&{\cfrac{d}{{dx}}(g(x))}&{\cfrac{d}{{dx}}(h(x))}\\l&m&n\\a&b&c\end{array}} \right|$ $+ \left| {\begin{array}{llllllllllllllllllll}{f(x)}&{g(x)}&{h(x)}\\0&0&0\\a&b&c\end{array}} \right| + \left| {\begin{array}{llllllllllllllllllll}{f(x)}&{g(x)}&{h(x)}\\l&m&n\\0&0&0\end{array}} \right|$
$= \left| {\begin{array}{llllllllllllllllllll}{f'(x)}&{g'(x)}&{h'(x)}\\l&m&n\\a&b&c\end{array}} \right|$
Hence proved.
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