For $$f(x)= \int_0^{\ln x} \frac{1}{\sqrt{4+e^t}}\, dt,\quad x>0,$$ find $(f^{-1})'(0)$
2026-03-29 03:36:52.1774755412
If $f(x)= \int_0^{\ln x} \frac{1}{\sqrt{4+e^t}},$ find $(f^{-1})'(0)$
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HINT : $$ f (x)=\int^{\ln x}_0\dfrac {1}{\sqrt {4+e^t}} dt $$ Thus, by the F.T.C, $$ f^{'}=\dfrac {1}{x\sqrt {4+x}}$$ Since $$(f^{-1})^{'}(x)=\dfrac {1}{f^{'}(f^{-1}(x))} $$ We have $$(f^{-1})^{'}(0)=x\sqrt {4+ f^{-1}(0)} $$