Question

Let $F\left(x\right)=f\left(x\right)\cdot g\left(x\right)\cdot h\left(x\right).$

At some point $x_0$ it is given that $F^{\prime }\left(x_0\right)=21F\left(x_0\right),$ $f^{\prime }\left(x_0\right)=4f\left(x_0\right),g^{\prime }\left(x_0\right)=-7g\left(x_0\right)$ and $h^{\prime }\left(x_0\right)=kh\left(x_0\right).$Then $k$ is

Answer 1

Given, $F\left(x\right)=f\left(x\right).g\left(x\right).h\left(x\right)$
On differentiating at $x=x_0,$ we get
$F^{\prime }\left(x_0\right)=f^{\prime }\left(x_0\right).g\left(x_0\right)h\left(x_0\right)+f\left(x_0\right).g^{\prime }\left(x_0\right)h\left(x_0\right)$
$+f\left(x_0\right)g\left(x_0\right)h^{\prime }\left(x_0\right)$ ....... $\left(i\right)$
where $F^{\prime }\left(x_0\right)=21F\left(x_0\right),f^{\prime }\left(x_0\right)$$=4f\left(x_0\right)$
$g^{\prime }\left(x_0\right)=-7g\left(x_0\right)$ and $h^{\prime }\left(x_0\right)=kh\left(x_0\right),$
On substituting in Eq. $\left(i\right),$ we get
$21F\left(x_0\right)=4f\left(x_0\right)g\left(x_0\right)h\left(x_0\right)-7f\left(x_0\right)g\left(x_0\right)h\left(x_0\right)+kf\left(x_0\right)g\left(x_0\right)h\left(x_0\right)$
$\Rightarrow 21=4-7+k$ [using $F\left(x_0\right)=f\left(x_0\right)g\left(x_0\right)h\left(x_0\right)$]
$\therefore k=24$