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Question: Let, f, g and h are differentiable functions, If\(f(0)=1\);\(g(0)=2\);\(h(0)=3\) and the derivatives...

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

Explanation

Solution

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)(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)=1f(0)=1 (fg)′(0)=6(fg)'(0)=6
g(0)=2g(0)=2 (gh)′(0)=4(gh)'(0)=4
h(0)=3h(0)=3 (hf)′(0)=5(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)(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)(fg)'(0)=f(0)g'(0)+f'(0)g(0)
Substitute the given values in above equation,
∴6=1×g′(0)+f′(0)×2\therefore 6=1\times g'(0)+f'(0)\times 2
∴g′(0)+2f′(0)=6\therefore g'(0)+2f'(0)=6…………………………………………….. (1)

(2). (gh)′(0)=g(0)h′(0)+g′(0)h(0)(gh)'(0)=g(0)h'(0)+g'(0)h(0)
Substitute the given values in above equation,
∴4=2×h′(0)+g′(0)×3\therefore 4=2\times h'(0)+g'(0)\times 3
∴2h′(0)+3g′(0)=4\therefore 2h'(0)+3g'(0)=4…………………………………………… (2)
(3). (hf)′(0)=h(0)f′(0)+h′(0)f(0)(hf)'(0)=h(0)f'(0)+h'(0)f(0)
Substitute the given values in above equation,
∴5=3×f′(0)+h′(0)×1\therefore 5=3\times f'(0)+h'(0)\times 1
∴3f′(0)+h′(0)=5\therefore 3f'(0)+h'(0)=5
∴h′(0)=5−3f′(0)\therefore h'(0)=5-3f'(0)…………………………………………… (3)
Put the value of equation (3) in equation (2),
∴2h′(0)+3g′(0)=4\therefore 2h'(0)+3g'(0)=4
∴2[5−3f′(0)]+3g′(0)=4\therefore 2\left[ 5-3f'(0) \right]+3g'(0)=4
∴10−6f′(0)+3g′(0)=4\therefore 10-6f'(0)+3g'(0)=4
∴3[−2f′(0)+g′(0)]=4−10\therefore 3\left[ -2f'(0)+g'(0) \right]=4-10
∴[−2f′(0)+g′(0)]=−63\therefore \left[ -2f'(0)+g'(0) \right]=\dfrac{-6}{3}
∴g′(0)−2f′(0)=−2\therefore g'(0)-2f'(0)=-2…………………………………………………… (4)
Now adding (1) and (2) we get,

& \text{ }g'(0)+2f'(0)=6 \\\ & \underline{+g'(0)-2f'(0)=-2} \\\ & 2g'(0)+0=4 \\\ \end{aligned}$$ $$\therefore g'(0)=\dfrac{4}{2}$$ $$\therefore g'(0)=2$$……………………………………………………………….. (5) $$\begin{aligned} & \therefore (fg)'(x)=16 \\\ & x=0 \\\ \end{aligned}$$ Put the value of (5) in (1), $$\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).