Question

$Ify=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|,provethat\frac{dy}{dx}=\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& m\\ a& b& c\end{array}\right|$

Answer 1

$Given,y=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|Expandalong{R}_1\Rightarrow y=(mc-nb)f\left(x\right)+(na-lc)g\left(x\right)+(lb-ma)h\left(x\right)Differentiatingw.r.t.x,\Rightarrow \frac{dy}{dx}=(mc-nb)f^{\prime }\left(x\right)+(na-lc)g^{\prime }\left(x\right)+(lb-ma)h^{\prime }\left(x\right)=\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& m\\ a& b& c\end{array}\right|$

Alternative to differentiate a determinant, we differentiate one row (on one column)at a time, keepping other unchange.

$y=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$

Differentiating w.r.t. x, we get

$\frac{dy}{dx}=\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& m\\ a& b& c\end{array}\right|+\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ 0& 0& 0\\ a& b& c\end{array}\right|+\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ 0& 0& 0\end{array}\right|=\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$