Question

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

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

Answer 1

To prove : $\frac{dy}{dx}=\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$
Expanding determinant along $R_1$

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

$\frac{dy}{dx}=(mc-bn)f^{\prime }\left(x\right)-(lc-an)g^{\prime }\left(x\right)+(lb-am)h^{\prime }\left(x\right)$

We need to prove that

$\frac{dy}{dx}=(mc-bn)f^{\prime }\left(x\right)-(lc-an)g^{\prime }\left(x\right)+(lb-am)h^{\prime }\left(x\right)$

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

Expanding determinant along $R_1$

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

$y=(mc-bn)f\left(x\right)-(lc-an)g\left(x\right)+(lb-am)h\left(x\right)$

Differentiating w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{d\left(\right(mc-bn\left)f\right(x)-(lc-an\left)g\right(x)+(lb-am\left)h\right(x\left)\right)}{dx}$

$\frac{dy}{dx}=\frac{d\left(\right(mc-bn\left)f\right(x\left)\right)}{dx}-\frac{d\left(\right(lc-an\left)g\right(x)}{dx}+\frac{d\left(\right(lb-am\left)h\right(x\left)\right)}{dx}$

$\frac{dy}{dx}=(mc-bn)\frac{d\left(f\right(x\left)\right)}{dx}-(lc-an)\frac{d\left(g\right(x)}{dx}+(lb-am)\frac{d\left(h\right(x\left)\right)}{dx}$

$\therefore \frac{dy}{dx}=(mc-bn)\left(f^{\prime }\right(x)-(lc-an\left)g^{\prime }\right(x)+(lb-am)+h^{\prime }(x)$

Hence, proved.