Find (f^-1)'(0)

Question

Find (f^-1)'(0), if f(x) = ∫₂^x (dt)/(sqrt(1+t^4)),f(2) = 0

Answer 1

f^-1(f(x)) = x

Thus, (f^-1(f(x)) )' = 1

Yet, (f^-1(f(x)) )' =(f^-1)^'(f(x)) f'(x), so (f^-1)^'(f(x)) = 1/ f'(x)

Then, from the fundamental theorem of calculus, f'(x) = 1/sqrt(1+x^4) and as f(2)=0, (f^-1)^'(0) = (f^-1)^'(f(2)) =

1/f'(2) = 1/(1/sqrt(1+2^4)) = sqrt(1+2^4) =sqrt(17)