The inverse function theorem says
[tex]\dfrac{\mathrm df^{-1}}{\mathrm dx}(16)=\dfrac1{\frac{\mathrm df}{\mathrm dx}(f^{-1}(16))}[/tex]
We have
[tex]f(x)=x^2+4x-5[/tex]
defined on [tex]x>-2[/tex], for which we get
[tex]f^{-1}(x)=-2+\sqrt{x+9}[/tex]
and
[tex]f^{-1}(16)=-2+\sqrt{16+9}=3[/tex]
The derivative of [tex]f(x)[/tex] is
[tex]f'(x)=2x+4[/tex]
So we end up with
[tex]\dfrac{\mathrm df^{-1}}{\mathrm dx}(16)=\dfrac1{\frac{\mathrm df}{\mathrm dx}(3)}=\dfrac1{10}[/tex]
