For x in R,let tan^{-1}(x) in left(-frac{pi}{ | vedclass

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(B) The function is given by $f(x) = \int_0^{x 	an^{-1} x} \frac{e^{t-\cos x}}{1+t^{2023}} dt$.Using Leibniz's rule to find the derivative $f'(x)$:$f

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$0$

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(B) The function is given by $f(x) = \int_0^{x 	an^{-1} x} \frac{e^{t-\cos x}}{1+t^{2023}} dt$.Using Leibniz's rule to find the derivative $f'(x)$:$f'(x) = \frac{e^{x 	an^{-1} x - \cos x}}{1 + (x 	an^{-1} x)^{2023}} \cdot \frac{d}{dx}(x 	an^{-1} x) + \int_0^{x 	an^{-1} x} \frac{\partial}{\partial x} \left( \frac{e^{t-\cos x}}{1+t^{2023}} ight) dt$.However,observing the structure of the function,we note that for $x=0$,$f(0) = \int_0^0 \dots dt = 0$.For $x 
eq 0$,$x 	an^{-1} x > 0$ because $	an^{-1} x$ has the same sign as $x$.Since the integrand $\frac{e^{t-\cos x}}{1+t^{2023}}$ is positive for $t \geq 0$,and the upper limit $x 	an^{-1} x$ is always non-negative,the integral $f(x)$ is non-negative for all $x \in R$.Specifically,$f(x) \geq 0$ for all $x$,and $f(0) = 0$.Thus,the minimum value of the function is $0$.

For $x \in R$,let $\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the minimum value of the function $f: R \rightarrow R$ defined by $f(x) = \int_0^{x \tan^{-1} x} \frac{e^{(t-\cos x)}}{1+t^{2023}} dt$ is

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