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(D) Given $f'(x) = \int_{0}^{f(x)} f^{-1}(t) dt$ and $f'(0) = 1$. Differentiating both sides with respect to $x$ using the Leibniz rule: $f''(x) = f^{

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$\sqrt{e}$

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(D) Given $f'(x) = \int_{0}^{f(x)} f^{-1}(t) dt$ and $f'(0) = 1$. Differentiating both sides with respect to $x$ using the Leibniz rule: $f''(x) = f^{-1}(f(x)) \cdot f'(x)$. Since $f(x)$ is invertible,$f^{-1}(f(x)) = x$. Thus,$f''(x) = x f'(x)$. Rearranging the terms,we get $\frac{f''(x)}{f'(x)} = x$. Integrating both sides with respect to $x$: $\ln|f'(x)| = \frac{x^2}{2} + C$. At $x = 0$,$f'(0) = 1$,so $\ln|1| = 0 + C$,which implies $C = 0$. Therefore,$\ln|f'(x)| = \frac{x^2}{2}$,which gives $f'(x) = e^{x^2/2}$. For $x = 1$,$f'(1) = e^{1/2} = \sqrt{e}$.

If $f(x)$ is an invertible and twice differentiable function satisfying $f'(x) = \int_{0}^{f(x)} f^{-1}(t) dt$ for all $x \in R$ and $f'(0) = 1$,then $f'(1)$ is equal to:

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