Question

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

Answer 1

Answer: 16

Let, $y=fgh$
$\frac{dy}{dx}=f^{\prime }gh+g^{\prime }h+fgh$
$=\frac{1}{2}\left(2f^{\prime }gh+2g^{\prime }h+2fg{h}^{\prime }\right)$
$=\frac{1}{2}\left(h\left(f^{\prime }g+g^{\prime }f\right)+g\left(f^{\prime }h+f{h}^{\prime }\right)+f\left(g^{\prime }h+g^{\prime }\right)\right)$
$=\frac{1}{2}\left[h.(fg{)}^{\prime }+g.(fh{)}^{\prime }+f.(gh{)}^{\prime }\right]$
$($ fgh) $(0)=\frac{1}{2}\left[h(0)(fg{)}^{\prime }(0)+g(0)(fh)(0)+f(0)(gh{)}^{\prime }(0)\right]$
$=\frac{1}{2}(3\times 6+2\times 5+1\times 4)$
$=\frac{1}{2}(18+10+4)=\frac{32}{2}=16$