Question

Lef $f,g$ and $h$ be differentiable functions. If $f\left(0\right)=1,$ $g\left(0\right)=2,$ $h\left(0\right)=3$ and the derivatives of their pairwise products at $x=0$ are $\left(fg{)}^{\prime }\right(0)=6,$ $\left(gh{)}^{\prime }\right(0)=4$ and $\left(hf{)}^{\prime }\right(0)=5,$ then the value of $\left(fgh{)}^{\prime }\right(0)$ is

Answer 1

Given that
$f\left(0\right)=1,$ $g\left(0\right)=2,$ $h\left(0\right)=3$ and
$\left(fg{)}^{\prime }\right(0)=6,(gh{)}^{\prime }\left(0\right)=4,\left(hf{)}^{\prime }\right(0)=5$

$f^{\prime }\left(0\right)g\left(0\right)+f\left(0\right)g^{\prime }\left(0\right)=6\Rightarrow 2f^{\prime }\left(0\right)+g^{\prime }\left(0\right)=6\dots \left(i\right)g^{\prime }\left(0\right)h\left(0\right)+g\left(0\right)h^{\prime }\left(0\right)=4\Rightarrow 3g^{\prime }\left(0\right)+2h^{\prime }\left(0\right)=4\dots \left(ii\right)h^{\prime }\left(0\right)f\left(0\right)+h\left(0\right)f^{\prime }\left(0\right)=5\Rightarrow h^{\prime }\left(0\right)+3f^{\prime }\left(0\right)=5\dots \left(iii\right)$

From $\left(i\right)$ and $\left(ii\right),$
$6f^{\prime }\left(0\right)-2h^{\prime }\left(0\right)=14\dots \left(iv\right)$
From $\left(iii\right)$ and $\left(iv\right),$
$f^{\prime }\left(0\right)=2,h^{\prime }\left(0\right)=-1$
Now, from equation $\left(i\right),$
$g^{\prime }\left(0\right)=2$

$\left(fgh{)}^{\prime }\right(0)=\sum f\left(0\right)g\left(0\right)h^{\prime }\left(0\right)=2\cdot 2\cdot 3+1\cdot 2\cdot 3+1\cdot 2\cdot (-1)=12+6-2=16$