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