## Calculus (3rd Edition)

$g'(x)= \frac{1}{7}.$
Since $f(x)=7x+6$, then $f'(x)=7$. Hence, by Theorem 2, and assuming that $g(x)=f^{-1}(x)$, we have $$g'(x)=\frac{1}{f'(x)}=\frac{1}{7}.$$