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 $\left(\text{fg}\right)^{'}\left(0\right)=6\text{;} \, \left(\text{gh}\right)^{'}\left(0\right)=4$ and $\left(\text{hf}\right)^{'}\left(0\right)=5$ , then the value of $\left(\text{fgh}\right)^{'}\left(0\right)$ is

Answer 1

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