Let e^{f(x)} = ln x. If g(x) is the inverse f | vedclass

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(C) Given $e^{f(x)} = \ln x$. Taking the natural logarithm on both sides,we get $f(x) = \ln(\ln x)$.Since $g(x)$ is the inverse function of $f(x)$,we

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$e^{(x + e^x)}$

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(C) Given $e^{f(x)} = \ln x$. Taking the natural logarithm on both sides,we get $f(x) = \ln(\ln x)$.Since $g(x)$ is the inverse function of $f(x)$,we have $f(g(x)) = x$.Substituting $g(x)$ into the expression for $f$,we get $\ln(\ln(g(x))) = x$.Exponentiating both sides,we get $\ln(g(x)) = e^x$.Exponentiating again,we get $g(x) = e^{e^x}$.Now,we differentiate $g(x)$ with respect to $x$ using the chain rule:$g'(x) = \frac{d}{dx}(e^{e^x}) = e^{e^x} \cdot \frac{d}{dx}(e^x) = e^{e^x} \cdot e^x$.Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get $g'(x) = e^{e^x + x}$.Thus,the correct option is $C$.

Let $e^{f(x)} = \ln x$. If $g(x)$ is the inverse function of $f(x)$,then $g'(x)$ is equal to:

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