Continuity and Differentiability — Class 12 Maths 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.