Answer
$g'(2)=\frac{1}{3} $
Work Step by Step
We have$f(g(x))=x$ because $g$ is the inverse of $f$.
Using the chain rule and the implicit differentiating it follows:
$$f(g(x))=x \to g'(x)f'(g(x))=(x)'$$
$$g'(x)f'(g(x))=1$$
$$g'(2)f'(g(2))=1$$
Since $f(g(x))=x$ it follows that $g(x)=f^{-1}(x)$.
So $g(2)=f^{-1}(2)=a$
$f^{-1}(2)=a \to 2=f(a) \to 2=2a+\ln(a) \to 2-2a=\ln(a) \to 2(1-a)=\ln(a) $
The equation $2(1-a)=\ln(a) $ has a trivial solution at $a=1$ because $2(1-1)=\ln(1) \to 2 \cdot 0=0 \to 0=0 $
Therefore, $g(2)=1=a$
$$g'(2)f'(g(2))=1 \to g'(2)f'(1)=1 $$
The first derivative of $f$ is:
$f'(x)=(2x)'+(\ln x)'=2+\frac{1}{x}$
$f'(1)=2+\frac{1}{1}=3$
So $$g'(2)f'(1)=1 \to g'(2)3=1 \to g'(2)=\frac{1}{3} $$
