Explanation
$$f(x)+f\left(\dfrac{1}{x}\right)=0$$ is satisfied by $$f(x)=k \ln x$$, where $$k$$ is a constant.
Given that $$f(e)=1 \Rightarrow k=1$$
$$f(x)= \ln x$$
$$ \Rightarrow f^{-1}x=e^x$$
$$ \Rightarrow g(x)=e^x$$
$$ \Rightarrow g'(x)=e^x$$
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