Question

If f, g and h are differentiable functions of x and $\Delta =\left|\begin{array}{ccc}f& g& h\\ (xf{)}^{\prime }& (xg{)}^{\prime }& (xh{)}^{\prime }\\ \left(x^{2}f\right)"& \left(x^{2}g\right)"& \left(x^{2}h\right)"\end{array}\right|$, then
${\Delta }^{\prime }=$

Answer 1

Given determinant may be expressed as
$\Delta =\left|\begin{array}{ccc}f& g& h\\ x{f}^{\prime }+f& x{g}^{\prime }+g& x{h}^{\prime }+h\\ (x^{2}f"+4x{f}^{\prime }+2f)& (x^{2}g"+4x{g}^{\prime }+2g)& (x^{2}h"+4x{h}^{\prime }+2h)\end{array}\right|$
$=\left|\begin{array}{ccc}f& g& h\\ x{f}^{\prime }& x{g}^{\prime }& x{h}^{\prime }\\ x^{2}f"& x^{2}g"& x^{2}h"\end{array}\right|R_3\to R_3-4R_2+2R_1;R_2\to R_2-R_1$
or $\Delta =x\left|\begin{array}{ccc}f& g& h\\ f^{\prime }& g^{\prime }& h^{\prime }\\ x^{2}f"& x^{2}g"& x^{2}h"\end{array}\right|$
$\Rightarrow \Delta =\left|\begin{array}{ccc}f& g& h\\ f^{\prime }& g^{\prime }& h^{\prime }\\ x^{3}f"& x^{3}g"& x^{3}h"\end{array}\right|$
$\therefore {\Delta }^{\prime }=\left|\begin{array}{ccc}{f}^{\prime }& g^{\prime }& h^{\prime }\\ f^{\prime }& g^{\prime }& h^{\prime }\\ x^{3}f"& x^{3}g"& x^{3}h"\end{array}\right|+\left|\begin{array}{ccc}f& g& h\\ f"& g"& h"\\ x^{3}f"& x^{3}g"& x^{3}h"\end{array}\right|+\left|\begin{array}{ccc}f& g& h\\ f^{\prime }& g^{\prime }& h^{\prime }\\ (x^{3}f"{)}^{\prime }& (x^{3}g"{)}^{\prime }& (x^{3}h"{)}^{\prime }\end{array}\right|$
$=0+0+\left|\begin{array}{ccc}f& g& h\\ f^{\prime }& g^{\prime }& h^{\prime }\\ (x^{3}f"{)}^{\prime }& (x^{3}g"{)}^{\prime }& (x^{3}h"{)}^{\prime }\end{array}\right|$
Hence
${\Delta }^{\prime }=\left|\begin{array}{ccc}f& g& h\\ f^{\prime }& g^{\prime }& h^{\prime }\\ (x^{3}f"{)}^{\prime }& (x^{3}g"{)}^{\prime }& (x^{3}h"{)}^{\prime }\end{array}\right|$