Suppose f(x) = frac{(2^x + 2^{-x}) tan x sqrt | vedclass

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(C) Given $f(x) = \frac{(2^x + 2^{-x}) 	an x \sqrt{	an^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$.First,calculate $f(0)$:$f(0) = \frac{(2^0 + 2^0) 	an

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$\sqrt{\frac{\pi}{4}}$

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(C) Given $f(x) = \frac{(2^x + 2^{-x}) 	an x \sqrt{	an^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$.First,calculate $f(0)$:$f(0) = \frac{(2^0 + 2^0) 	an(0) \sqrt{	an^{-1}(0-0+1)}}{(0+0+1)^3} = \frac{2 	imes 0 	imes \sqrt{\pi/4}}{1} = 0$.Now,use the definition of the derivative at $x=0$:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{f(h)}{h}$.$f'(0) = \lim_{h 	o 0} \left[ \frac{2^h + 2^{-h}}{(7h^2 + 3h + 1)^3} \cdot \frac{	an h}{h} \cdot \sqrt{	an^{-1}(h^2 - h + 1)} ight]$.As $h 	o 0$:$\frac{2^h + 2^{-h}}{(7h^2 + 3h + 1)^3} 	o \frac{1+1}{1} = 2$.$\frac{	an h}{h} 	o 1$.$\sqrt{	an^{-1}(h^2 - h + 1)} 	o \sqrt{	an^{-1}(1)} = \sqrt{\frac{\pi}{4}}$.Therefore,$f'(0) = 2 	imes 1 	imes \sqrt{\frac{\pi}{4}} = 2 	imes \frac{\sqrt{\pi}}{2} = \sqrt{\pi}$.

Suppose $f(x) = \frac{(2^x + 2^{-x}) \tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$. Then the value of $f'(0)$ is equal to

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