Find (f^-1)'(a): f(x) = 3+x^2+tan(pi(x)/2), -1

(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