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# Let $f$, $g$ and $h$ are differentiable functions. If $f\left( 0 \right) = 1$; $g\left( 0 \right) = 2$; $h\left( 0 \right) = 3$ and the derivative of their pairwise products at $x = 0$ are $\left( {fg} \right)'\left( 0 \right) = 6$; $\left( {gh} \right)'\left( 0 \right) = 4$ and $\left( {hf} \right)'\left( 0 \right) = 5$ then compute the value of $\left( {fgh} \right)'\left( 0 \right)$.A.12B.15C.16D.None of these  Verified
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Hint: Here we will first differentiate the function $fgh$ using the formula of differentiation. Then we will modify the equation such that we get the equation in terms of the values given in the question. Then we will put their values in the equation and solve it to get the final value of $\left( {fgh} \right)'\left( 0 \right)$.

Formula used:
We will use the following formulas:
1. $\dfrac{d}{{dx}}\left( {uv} \right) = uv' + u'v$
2.$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}$

Let $y = fgh$.
Now we will differentiate the function with respect to $x$. Therefore, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {fgh} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = f'gh + fg'h + fgh'$
We will write the above equation in modified form to solve the equation with the given values. Therefore, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {2f'gh + 2fg'h + 2fgh'} \right)$
We will write the above equation as
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {h\left( {f'g + fg'} \right) + g\left( {f'h + fh'} \right) + f\left( {g'h + gh'} \right)} \right)$
Using the differentiation property, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \left( {fgh} \right)' = \dfrac{1}{2}\left( {h\left( {fg} \right)' + g\left( {fh} \right)' + f\left( {gh} \right)'} \right)$
Now we have to find the value of $\left( {fgh} \right)'\left( 0 \right)$.
Therefore, we get
$\Rightarrow \left( {fgh} \right)'\left( 0 \right) = \dfrac{1}{2}\left( {h\left( 0 \right)\left( {fg} \right)'\left( 0 \right) + g\left( 0 \right)\left( {fh} \right)'\left( 0 \right) + f\left( 0 \right)\left( {gh} \right)'\left( 0 \right)} \right)$
Substituting $f\left( 0 \right) = 1$, $g\left( 0 \right) = 2$, $h\left( 0 \right) = 3$, $\left( {fg} \right)'\left( 0 \right) = 6$, $\left( {gh} \right)'\left( 0 \right) = 4$ and $\left( {hf} \right)'\left( 0 \right) = 5$ in the above equation, we get
$\Rightarrow \left( {fgh} \right)'\left( 0 \right) = \dfrac{1}{2}\left( {3 \times 6 + 2 \times 5 + 1 \times 4} \right)$
Multiplying the terms, we get
$\Rightarrow \left( {fgh} \right)'\left( 0 \right) = \dfrac{1}{2}\left( {18 + 10 + 4} \right)$
$\Rightarrow \left( {fgh} \right)'\left( 0 \right) = \dfrac{{32}}{2}$
$\Rightarrow \left( {fgh} \right)'\left( 0 \right) = 16$
Hence, the value of $\left( {fgh} \right)'\left( 0 \right)$ is equal to 16.