If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)

If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)

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\[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\]
find \[(f ^{-1})'(0)\]

f'(x) = sqrt(1+x^3)
so f'(0= 1

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