Let f(x) = intlimits_2^x frac{dt}{sqrt{1 + t^ | vedclass

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(C) Given $f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}}$.By the Fundamental Theorem of Calculus,$f'(x) = \frac{1}{\sqrt{1 + x^4}}$.Since $g$ is th

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$\sqrt{17}$

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(C) Given $f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}}$.By the Fundamental Theorem of Calculus,$f'(x) = \frac{1}{\sqrt{1 + x^4}}$.Since $g$ is the inverse of $f$,we have $g(f(x)) = x$. Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)} = \sqrt{1 + x^4}$.To find $g'(0)$,we need to find $x$ such that $f(x) = 0$.$f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}} = 0$ implies $x = 2$.Therefore,$g'(0) = g'(f(2)) = \sqrt{1 + 2^4} = \sqrt{1 + 16} = \sqrt{17}$.

Let $f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}}$ and $g$ be the inverse of $f$. Then the value of $g'(0)$ is

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