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Step-by-Step Solution
Step 1: Break down the given function
We have
$f(x) = \frac{\left(2^x + 2^{-x}\right)\,\tan x \,\sqrt{\tan^{-1}\bigl(x^2 - x + 1\bigr)}}{\left(7 x^2 + 3 x + 1\right)^3}.$
To simplify the differentiation, define the sub-functions:
$g(x) = 2^x + 2^{-x}$
$h(x) = \tan x$
$i(x) = \sqrt{\tan^{-1}(x^2 - x + 1)}$
$j(x) = \bigl(7 x^2 + 3 x + 1\bigr)^3$
Thus,
$f(x) = \frac{g(x)\,h(x)\,i(x)}{j(x)}.$
Step 2: Compute the values and derivatives of each part at $x = 0$
$g(x) = 2^x + 2^{-x}$
• At $x = 0:$
$g(0) = 2^0 + 2^0 = 2.$
• Derivative:
$g'(x) = 2^x \ln 2 - 2^{-x} \ln 2.$
So
$g'(0) = \ln 2 - \ln 2 = 0.$
$h(x) = \tan x$
• At $x = 0:$
$h(0) = \tan(0) = 0.$
• Derivative:
$h'(x) = \sec^2 x,$
hence
$h'(0) = \sec^2(0) = 1.$
$i(x) = \sqrt{\tan^{-1}(x^2 - x + 1)}$
Let $u(x) = \tan^{-1}(x^2 - x + 1).$ Then $i(x) = \sqrt{u(x)}.$
• At $x = 0:$
$u(0) = \tan^{-1}(1) = \frac{\pi}{4},$
so
$i(0) = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}.$
• Next, we find $u'(x)$:
$$
u'(x) = \frac{2x - 1}{1 + (x^2 - x + 1)^2}.
$$
Thus
$u'(0) = \frac{-1}{1 + (1)^2} = \frac{-1}{2}.$
• By the chain rule for $i(x) = \sqrt{u(x)}$:
$$
i'(x) = \frac{1}{2\sqrt{u(x)}} \, u'(x).
$$
So
$$
i'(0)
= \frac{1}{2 \cdot \bigl(\sqrt{\pi}/2\bigr)} \cdot \left(-\frac{1}{2}\right)
= \frac{1}{\sqrt{\pi}} \cdot \left(-\frac{1}{2}\right)
= -\frac{1}{2\sqrt{\pi}}.
$$
$j(x) = \bigl(7x^2 + 3x + 1\bigr)^3$
Let $k(x) = 7x^2 + 3x + 1.$ Then $j(x) = \bigl(k(x)\bigr)^3.$
• Derivative of $k(x)$:
$k'(x) = 14x + 3.$
Hence,
$$
j'(x) = 3 \,\bigl(k(x)\bigr)^2 \cdot k'(x).
$$
• At $x = 0:$
$k(0) = 1,$
$k'(0) = 3,$
so
$j(0) = 1^3 = 1,$
and
$j'(0) = 3 \times 1^2 \times 3 = 9.$
Step 3: Apply the quotient rule for $f(x) = \frac{g(x)\,h(x)\,i(x)}{j(x)}$
The derivative is:
$$
f'(x)
= \frac{\bigl[g'(x)\,h(x)\,i(x) + g(x)\,h'(x)\,i(x) + g(x)\,h(x)\,i'(x)\bigr]\;j(x)\;-\;g(x)\,h(x)\,i(x)\,j'(x)}{\bigl(j(x)\bigr)^2}.
$$
Step 4: Evaluate at $x = 0$
Substitute all known values at $x=0$:
$g(0) = 2,$ $g'(0) = 0,$
$h(0) = 0,$ $h'(0) = 1,$
$i(0) = \frac{\sqrt{\pi}}{2},$ $i'(0) = -\frac{1}{2\sqrt{\pi}},$
$j(0) = 1,$ $j'(0) = 9.$
Numerator at $x=0$:
$$
\bigl[\,g'(0)\,h(0)\,i(0) \;+\; g(0)\,h'(0)\,i(0) \;+\; g(0)\,h(0)\,i'(0)\bigr]\;j(0)\;-\;g(0)\,h(0)\,i(0)\,j'(0).
$$
Calculate term by term:
$g'(0)\,h(0)\,i(0) = 0 \times 0 \times \frac{\sqrt{\pi}}{2} = 0,$
$g(0)\,h'(0)\,i(0) = 2 \times 1 \times \frac{\sqrt{\pi}}{2} = \sqrt{\pi},$
$g(0)\,h(0)\,i'(0) = 2 \times 0 \times \bigl(-\frac{1}{2\sqrt{\pi}}\bigr) = 0.$
So their sum is $\sqrt{\pi}$. Multiplying by $j(0)=1$ leaves it as $\sqrt{\pi}$. Subtracting the term
$g(0)\,h(0)\,i(0)\,j'(0) = 2 \times 0 \times \frac{\sqrt{\pi}}{2} \times 9 = 0,$
the numerator becomes $\sqrt{\pi}.$
The denominator at $x=0$ is $[j(0)]^2 = 1^2 = 1.$
Step 5: Conclude the value of $f'(0)$
Therefore,
$$
f'(0) = \frac{\sqrt{\pi}}{1} = \sqrt{\pi}.
$$
Hence, the correct value is $\sqrt{\pi}.$
