Question

If $y=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ 1& m& n\\ a& b& c\end{array}\right|$, then $\frac{dy}{dx}$ is equal to

• A. $\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$
• B. $\left|\begin{array}{ccc}l& m& n\\ f^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ a& b& c\end{array}\right|$
• C. $\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& l& a\\ g^{\prime }\left(x\right)& m& b\\ h^{\prime }\left(x\right)& n& c\end{array}\right|$
• D. $\left|\begin{array}{ccc}l& m& n\\ a& b& c\\ f^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\end{array}\right|$

Answer 1

Answer: (A), (C), (D)

We have,
$y=\left[\begin{array}{ccc}f(x)& g(x)& h(x)\\ 1& m& n\\ a& b& c\end{array}\right]$
$\therefore \frac{dy}{dx}=\left[\begin{array}{ccc}\frac{d}{dx}f(x)& \frac{d}{dx}g(x)& \frac{d}{dx}h(x)\\ l& m& n\\ a& b& c\end{array}\right]$
$+\left[\begin{array}{ccc}f(x)& g(x)& h(x)\\ \frac{d}{dx}& \frac{d}{dx}m& \frac{d}{dx}\\ a& b& c\end{array}\right]+\left[\begin{array}{ccc}f(x)& g(x)& h(x)\\ 1& m& n\\ \frac{d}{dx}a& \frac{d}{dx}b& \frac{d}{dx}c\end{array}\right]$
$=\left[\begin{array}{ccc}{f}^{\prime }(x)& g^{\prime }(x)& h^{\prime }(x)\\ l& m& n\\ a& b& c\end{array}\right]+\left[\begin{array}{ccc}f(x)& g(x)& h(x)\\ 0& 0& 0\\ a& b& c\end{array}\right]$
$+\left[\begin{array}{ccc}f(x)& g(x)& h(x)\\ l& m& n\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}{f}^{\prime }(x)& g^{\prime }(x)& h^{\prime }(x)\\ l& m& n\\ a& b& c\end{array}\right]$
$=\left[\begin{array}{ccc}& m& n\\ a& b& c\\ f^{\prime }(x)& g^{\prime }(x)& h^{\prime }(x)\end{array}\right]=\left[\begin{array}{ccc}{f}^{\prime }(x)& 1& a\\ g^{\prime }(x)& m& b\\ h^{\prime }(x)& n& c\end{array}\right]$