Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^2} + 1;\,\,\,\,\,\left( {5,2} \right) \cr
& {\text{Calculate }}f'\left( 2 \right) \cr
& f'\left( x \right) = 2x \cr
& f'\left( 2 \right) = 2\left( 2 \right) \cr
& f'\left( 2 \right) = 4 \cr
& {\text{Find the derivative of the inverse function}}{\text{, using the THEOREM 3}}{\text{.23}} \cr
& \left( {{f^{ - 1}}} \right)'\left( {{y_0}} \right) = \frac{1}{{f'\left( {{x_0}} \right)}},\,\,\,\,{\text{Where }}{y_0} = f\left( {{x_0}} \right) \cr
& {\text{Then}}{\text{,}} \cr
& \left( {{f^{ - 1}}} \right)'\left( 5 \right) = \frac{1}{{f'\left( 2 \right)}} \cr
& \left( {{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{4} \cr} $$
