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(A) Given that $f(x)$ and $g(x)$ are inverse functions, we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$f'(

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$\frac{1}{1 + e^{g(x)}}$

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(A) Given that $f(x)$ and $g(x)$ are inverse functions, we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$f'(g(x)) \cdot g'(x) = 1$.Therefore, $g'(x) = \frac{1}{f'(g(x))}$.Given $f(x) = x + e^x$, we find the derivative $f'(x) = 1 + e^x$.Substituting $g(x)$ into $f'(x)$, we get $f'(g(x)) = 1 + e^{g(x)}$.Thus, $g'(x) = \frac{1}{1 + e^{g(x)}}$.

If $f(x)$ and $g(x)$ are inverse functions of each other and $f(x) = x + e^x$, then $g'(x) = ...$

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