We can use the inverse function derivative theorem:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.[/tex]
In this case, we want to evaluate [tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}[/tex], so:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.[/tex]
The derivative is:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.[/tex]
The ordinate of the point is [tex]f^{-1}(1) = 0[/tex], so we evaluate:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.[/tex]
Finally:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.[/tex]
We can check the answer by finding the inverse:
[tex]y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},[/tex]
so that
[tex]f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.[/tex]
Therefore:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.[/tex]
Which finally gives the same answer as before:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.[/tex]
Answer: [tex]\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.[/tex]
