Question

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 $(fg)'(0)=6$;$(gh)'(0)=4$;$(hf)'(0)=5$then compute the value of $(fgh)'(0)$.

Answer 1

Hint: Use the formula given below to solve the problem.
\[(fgh)'(x)=f(x)g(x)h'(x)+f(x)g'(x)h(x)+f'(x)g(x)h(x)\]

First we will write the given values,
$f(0)=1$ $(fg)'(0)=6$
$g(0)=2$ $(gh)'(0)=4$
$h(0)=3$ $(hf)'(0)=5$
To proceed further we should know the formula given below,

Formula:
If f and g are two different functions then,
$(fg)'(x)=f(x)g'(x)+f'(x)g(x)$ ( Formulated by simply using chain rule.)
By using above formula we can write,

Complete step-by-step solution -

(1). $(fg)'(0)=f(0)g'(0)+f'(0)g(0)$
Substitute the given values in above equation,
\[\therefore 6=1\times g'(0)+f'(0)\times 2\]
\[\therefore g'(0)+2f'(0)=6\]…………………………………………….. (1)

(2). \[(gh)'(0)=g(0)h'(0)+g'(0)h(0)\]
Substitute the given values in above equation,
\[\therefore 4=2\times h'(0)+g'(0)\times 3\]
\[\therefore 2h'(0)+3g'(0)=4\]…………………………………………… (2)
(3). \[(hf)'(0)=h(0)f'(0)+h'(0)f(0)\]
Substitute the given values in above equation,
\[\therefore 5=3\times f'(0)+h'(0)\times 1\]
\[\therefore 3f'(0)+h'(0)=5\]
\[\therefore h'(0)=5-3f'(0)\]…………………………………………… (3)
Put the value of equation (3) in equation (2),
\[\therefore 2h'(0)+3g'(0)=4\]
\[\therefore 2\left[ 5-3f'(0) \right]+3g'(0)=4\]
\[\therefore 10-6f'(0)+3g'(0)=4\]
\[\therefore 3\left[ -2f'(0)+g'(0) \right]=4-10\]
\[\therefore \left[ -2f'(0)+g'(0) \right]=\dfrac{-6}{3}\]
\[\therefore g'(0)-2f'(0)=-2\]…………………………………………………… (4)
Now adding (1) and (2) we get,
\[\begin{align}
  & \text{ }g'(0)+2f'(0)=6 \\
 & \underline{+g'(0)-2f'(0)=-2} \\
 & 2g'(0)+0=4 \\
\end{align}\]
\[\therefore g'(0)=\dfrac{4}{2}\]
\[\therefore g'(0)=2\]……………………………………………………………….. (5)
\[\begin{align}
  & \therefore (fg)'(x)=16 \\
 & x=0 \\
\end{align}\]
Put the value of (5) in (1),
\[\therefore g'(0)+2f'(0)=6\]
\[\therefore 2+2f'(0)=6\]
\[\therefore f'(0)=\dfrac{6-2}{2}\]
\[\therefore f'(0)=\dfrac{4}{2}\]
\[\therefore f'(0)=2\]……………………………………………………… (6)
Also put the value of (5) in (2),
\[\therefore 2h'(0)+3g'(0)=4\]
\[\therefore 2h'(0)+3\times 2=4\]
\[\therefore 2h'(0)+6=4\]
\[\therefore 2h'(0)=4-6\]
\[\therefore h'(0)=\dfrac{-2}{2}\]
\[\therefore h'(0)=-1\]……………………………………………………. (7)
To proceed further we should know the formula given below,
Formula:
If f and g are two different functions then,
\[(fgh)'(x)=f(x)g(x)h'(x)+f(x)g'(x)h(x)+f'(x)g(x)h(x)\] (Formulated by simply using chain rule.)
If we put \[x=0\] then,
\[(fgh)'(0)=f(0)g(0)h'(0)+f(0)g'(0)h(0)+f'(0)g(0)h(0)\]
Now, substitute the values from given equations and from equations (5), (6), (7) in above equation,
\[\therefore (fgh)'(0)=1\times 2\times (-1)+1\times 2\times 3+2\times 2\times 3\]
\[\therefore (fgh)'(0)=-2+6+12\]
\[\therefore (fgh)'(0)=-2+18\]
\[\therefore (fgh)'(0)=16\]
Therefore the value of \[(fgh)'(0)\] is \[16\]

Note: Here is in fact what will confuse all of us is the formula of \[(fg)'(x)\]which is evaluated simply using chain rule as it is a multiplication of function. Some will assume it as a composite function but it is not as composite functions have proper notations like, f [g(x)] or (fog)(x).