Question

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

Answer 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