Let f, g and h are differentiable functions. If f(0) = 1; g(0) = 2; h(0) = 3 and the derivatives of their pair wise products at x = 0 are (fg) '(0) = 6;(gh) '(0) = 4 and (hf) ' (0) = 5 , then the value of (fgh)'(0) & is

Correct answer: 16 Let, y = fgh dy dx = f ^ ' gh + fg ^ ' h + fgh ^ ' = (1)/(2)(2 f ^ ' gh + 2 fg ^ ' h + 2 fgh ^ ' ) = (1)/(2)(h( f ^ ' g + g ^ ' f) + g( f ^ ' h + fh ^ ' ) + f( g ^ ' h + gh ^ ' )) = (1)/(2)[h. (fg) ^ ' + g. (fh) ^ ' + f. (gh) ^ ' ] (fgh) ^ ' (0) = (1)/(2)[h(0) (fg) ^ ' (0) + g(0) (fh) ^ ' (0) + f(0) (gh) ^ ' (0)] = (1)/(2) (3 × 6 + 2 × 5 + 1 × 4) = (1)/(2) (1 8 + 1 0 + 4) = (3 2)/(2) = 1 6

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

Let f, g and h are differentiable functions. If f(0) = 1; g(0) = 2;...

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

Correct answer

Step-by-step solution

Let,

\text{y} = \text{fgh}

\dfrac{\text{dy}}{\text{dx}} = {\text{f}}^{'}\text{gh} + {\text{fg}}^{'}\text{h} + {\text{fgh}}^{'}

= \dfrac{1}{2}\left(2{\text{f}}^{'}\text{gh} + 2{\text{fg}}^{'}\text{h} + 2{\text{fgh}}^{'}\right)

= \dfrac{1}{2}\left(\text{h}\left({\text{f}}^{'}\text{g} + {\text{g}}^{'}\text{f}\right) + \text{g}\left({\text{f}}^{'}\text{h} + {\text{fh}}^{'}\right) + \text{f}\left({\text{g}}^{'}\text{h} + {\text{gh}}^{'}\right)\right)

= \dfrac{1}{2}\left[\text{h}\text{.}{\left(\text{fg}\right)}^{'} + \text{g}\text{.}{\left(\text{fh}\right)}^{'} + \text{f}\text{.}{\left(\text{gh}\right)}^{'}\right]

{\left(\text{fgh}\right)}^{'}\left(0\right) = \dfrac{1}{2}\left[\text{h}\left(0\right){\left(\text{fg}\right)}^{'}\left(0\right) + \text{g}\left(0\right){\left(\text{fh}\right)}^{'}\left(0\right) + \text{f}\left(0\right){\left(\text{gh}\right)}^{'}\left(0\right)\right]

= \dfrac{1}{2} \left(3 \times 6 + 2 \times 5 + 1 \times 4\right)

= \dfrac{1}{2} \left(1 8 + 1 0 + 4\right) = \dfrac{3 2}{2} = 1 6

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