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Question Number 132076 by liberty last updated on 11/Feb/21
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 )) .
$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{−\mathrm{1}/\mathrm{2}} {dt}. \\ $$$$\mathrm{If}\:\mathrm{h}\left({x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{f}\left({x}\right)\:\mathrm{and}\:\mathrm{h}'\left({x}\right) \\ $$$$\mathrm{is}\:\mathrm{derivative}\:\mathrm{of}\:\mathrm{h}\left({x}\right).\:\mathrm{Find}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{h}''\left({x}\right)}{\left(\mathrm{h}\left(\mathrm{x}\right)\right)^{\mathrm{2}} }\:. \\ $$$$ \\ $$
Answered by EDWIN88 last updated on 11/Feb/21
We have f(x)=∫_0 ^( x) (1+t^3 )^(−1/2) dt , then f(g(x))= ∫_0 ^(g(x)) (1+t^3 )^(−1/2) dt x = ∫_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 ⇒(g′(x))^2 =1+g(x)^3 differentiating again 2g′(x).g′′(x)=3g(x)^2 .g′(x)⇒g′′(x)=(3/2)(g(x))^2 thus we find ((g′′(x))/(g^2 (x))) = (3/2)
$$\mathrm{We}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{x}} \left(\mathrm{1}+\mathrm{t}^{\mathrm{3}} \right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{dt}\:,\:\mathrm{then} \\ $$$$\:\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\:\int_{\mathrm{0}} ^{\mathrm{g}\left(\mathrm{x}\right)} \left(\mathrm{1}+\mathrm{t}^{\mathrm{3}} \right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{dt} \\ $$$$\:\mathrm{x}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{g}\left(\mathrm{x}\right)} \left(\mathrm{1}+\mathrm{t}^{\mathrm{3}} \right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{dt}\: \\ $$$$\:\mathrm{differentiating}\:\mathrm{both}\:\mathrm{sides}\: \\ $$$$\:\mathrm{1}\:=\:\left(\mathrm{1}+\left(\mathrm{g}\left(\mathrm{x}\right)^{\mathrm{3}} \right)^{−\mathrm{1}/\mathrm{2}} \mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides}\right. \\ $$$$\mathrm{1}=\left(\mathrm{1}+\mathrm{g}\left(\mathrm{x}\right)^{\mathrm{3}} \right)^{−\mathrm{1}} \left(\mathrm{g}'\left(\mathrm{x}\right)\right)^{\mathrm{2}} \Rightarrow\left(\mathrm{g}'\left(\mathrm{x}\right)\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{g}\left(\mathrm{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{differentiating}\:\mathrm{again}\: \\ $$$$\mathrm{2g}'\left(\mathrm{x}\right).\mathrm{g}''\left(\mathrm{x}\right)=\mathrm{3g}\left(\mathrm{x}\right)^{\mathrm{2}} .\mathrm{g}'\left(\mathrm{x}\right)\Rightarrow\mathrm{g}''\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{g}\left(\mathrm{x}\right)\right)^{\mathrm{2}} \\ $$$$\mathrm{thus}\:\mathrm{we}\:\mathrm{find}\:\frac{\mathrm{g}''\left(\mathrm{x}\right)}{\mathrm{g}^{\mathrm{2}} \left(\mathrm{x}\right)}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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