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-

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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