Questions & Answers

Question

Answers

A.12

B.15

C.16

D.None of these

Answer

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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

\[ \Rightarrow \left( {fgh} \right)'\left( 0 \right) = \dfrac{1}{2}\left( {18 + 10 + 4} \right)\]

Adding the terms, we get

\[ \Rightarrow \left( {fgh} \right)'\left( 0 \right) = \dfrac{{32}}{2}\]

Dividing 32 by 2, we get

\[ \Rightarrow \left( {fgh} \right)'\left( 0 \right) = 16\]

Hence, the value of \[\left( {fgh} \right)'\left( 0 \right)\] is equal to 16.

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