Respuesta :

(f^-1)'(a)=1/f'(f^-1)'(a)
f(a) =3 = 3+x^2+tan(pi(x)/2), 
0= 3+x^2+tan(pi(x)/2)
(f^-1)'(a)
= 1/[f ' ( f^-1(a) )] (f^-1)'(3)
 = 1/[f ' ( f^-1(3) )] (f^-1)'(3)
 = 1/[f ' ( 0 )] (f^-1)'(3)
=0+1/2pie(1)
=pi/2
hope it helps

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