Given the function f(x)=∫_0 ^( x) (1+t^3 )^(−1/2) dt. If h(x) is the inverse of f(x) and h′(x) is derivative of h(x). Find the value of ((h′′(x))/((h(x))^2 )) .


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Question Number 132076 by liberty last updated on 11/Feb/21

$Given the function f (x) = \int_{0}^{x} {(1 + t^{3})}^{- 1 / 2} d t .$ $If h (x) is the inverse of f (x) and h^{'} (x)$ $is derivative of h (x) . Find the$ $value of \frac{h^{″} (x)}{{(h (x))}^{2}} .$
	Answered by EDWIN88 last updated on 11/Feb/21

	$We have f (x) = \int_{0}^{x} {(1 + t^{3})}^{- 1 / 2} dt, then$ $f (g (x)) = \int_{0}^{g (x)} {(1 + t^{3})}^{- 1 / 2} dt$ $x = \int_{0}^{g (x)} {(1 + t^{3})}^{- 1 / 2} dt$ $differentiating both sides$ $1 = (1 + {(g {(x)}^{3})}^{- 1 / 2} g^{'} (x) squaring both sides$ $1 = {(1 + g {(x)}^{3})}^{- 1} {(g^{'} (x))}^{2} \Rightarrow {(g^{'} (x))}^{2} = 1 + g {(x)}^{3}$ $differentiating again$ ${2 g}^{'} (x) . g^{″} (x) = 3 g {(x)}^{2} . g^{'} (x) \Rightarrow g^{″} (x) = \frac{3}{2} {(g (x))}^{2}$ $thus we find \frac{g^{″} (x)}{g^{2} (x)} = \frac{3}{2}$
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