Answer
$\frac{1}{2}$
Work Step by Step
Using the definition of the inverse of the derivative of a function it follows that:
$$(f^{-1})'(4)=\frac{1}{f'(a)}~~\text{where}~~f(a)=4$$
Find $a$.
$$f(a)=4 \to 3+a+e^{a}=4 \to a+e^{a}=1$$
The trivial solution of the above equation is $a=0$ because $0+e^{0}=0+1=1$
$$(f^{-1})'(4)=\frac{1}{f'(a)} \to (f^{-1})'(4)=\frac{1}{f'(0)}$$
$$f(x)=3+x+e^{x} \to f'(x)=(3)'+(x)'+(e^{x})' \to f'(x)=1+e^{x} \to f'(0)=1+e^{0}=1+1=2$$
$$(f^{-1})'(4)=\frac{1}{f'(0)} \to (f^{-1})'(4)=\frac{1}{2} $$
