Answer
$\frac{1}{2}$
Work Step by Step
Given: $f(x)=x+e^x$
To find $(f^{-1})'(1)$, we need to find $f^{-1}(1)$ and $f'(f^{-1}(1))$.
Find $f^{-1}(1)$:
$f(0)=0+e^0$
$f(0)=0+1$
$f(0)=1$
$f^{-1}(1)=0$
Find $f'(0)$:
$f'(x)=\frac{d}{dx}(x+e^x)$
$f'(x)=1+e^x$
$f'(0)=1+e^0$
$f'(0)=1+1$
$f'(0)=2$
Using the Formula in Exercise 83,
$(f^{-1})'(1)=\frac{1}{f'(f^{-1}(1))}=\frac{1}{f'(0)}=\frac{1}{2}$
