If a function f(x) has an inverse f ⁻¹(x), then by definition
[tex]f\left(f^{-1}(x)\right) = x[/tex]
Differentiating both sides with respect to x yields
[tex]f'\left(f^{-1}(x)\right) \times \left(f^{-1}\right)'(x) = 1 \implies \left(f^{-1}\right)'(x) = \dfrac1{f'\left(f^{-1}(x)\right)}[/tex]
Now, if f(a) = b and b = f ⁻¹(a), then
[tex]\left(f^{-1}\right)'(a) = \dfrac1{f'\left(f^{-1}(a)\right)} = \dfrac1{f'(b)}[/tex]
