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Question Number 220224 by Mamadi last updated on 09/May/25
calcul together of definition of and calcul the derive f^ ^′ f(x)= x(√((x−1)/(x+1)))
$${calcul}\:{together}\:{of}\:{definition}\:{of}\:{and} \\ $$$${calcul}\:{the}\:{derive}\:{f}^{\:} \:^{'} \\ $$$${f}\left({x}\right)=\:\:{x}\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$
Answered by hlwrc last updated on 09/May/25
((df(x))/dx)=(√((x−1)/(x+1)))+(x/2)(√((x+1)/(x−1)))×(((x−1)−(x+1))/((x+1)^2 )) ==(√((x−1)/(x+1)))−((2x)/(2(x+1)(√((x+1)(x−1)))))
$$\frac{{df}\left({x}\right)}{{dx}}=\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}+\frac{{x}}{\mathrm{2}}\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}×\frac{\left({x}−\mathrm{1}\right)−\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$==\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}−\frac{\mathrm{2}{x}}{\mathrm{2}\left({x}+\mathrm{1}\right)\sqrt{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by efronzo1 last updated on 09/May/25
f(x) =(√((x^3 −x^2 )/(x+1))) f ′(x) = ((((3x^2 −2x)(x+1)−(x^3 −x^2 ))/((x+1)^2 ))/(2(√((x^3 −x^2 )/(x+1)))))
$$\:\mathrm{f}\left(\mathrm{x}\right)\:=\sqrt{\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{1}}}\: \\ $$$$\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\frac{\left(\mathrm{3x}^{\mathrm{2}} −\mathrm{2x}\right)\left(\mathrm{x}+\mathrm{1}\right)−\left(\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{1}}}} \\ $$
Answered by Frix last updated on 10/May/25
f(x)=((x(√(x−1)))/( (√(x+1))))=(u/v) ⇒ f′(x)=((u′v−v′u)/v^2 ) u=gh ⇒ u′=g′h+h′g g=x ⇒ g′=1 h=(√(x−1)) ⇒ h′=(1/(2(√(x−1)))) v=(√(x+1)) ⇒ v′=(1/(2(√(x+1)))) f′(x)=(((g′h+h′g)v−v′gh)/v^2 )= =((((√(x−1))+(x/(2(√(x−1)))))(√(x+1))−((x(√(x−1)))/(2(√(x+1)))))/(x+1))= ... =((x^2 +x−1)/( (√((x+1)^3 (x−1)))))
$${f}\left({x}\right)=\frac{{x}\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}}=\frac{{u}}{{v}}\:\Rightarrow\:{f}'\left({x}\right)=\frac{{u}'{v}−{v}'{u}}{{v}^{\mathrm{2}} } \\ $$$${u}={gh}\:\Rightarrow\:{u}'={g}'{h}+{h}'{g} \\ $$$${g}={x}\:\Rightarrow\:{g}'=\mathrm{1} \\ $$$${h}=\sqrt{{x}−\mathrm{1}}\:\Rightarrow\:{h}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$${v}=\sqrt{{x}+\mathrm{1}}\:\Rightarrow\:{v}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}} \\ $$$${f}'\left({x}\right)=\frac{\left({g}'{h}+{h}'{g}\right){v}−{v}'{gh}}{{v}^{\mathrm{2}} }= \\ $$$$=\frac{\left(\sqrt{{x}−\mathrm{1}}+\frac{{x}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}}\right)\sqrt{{x}+\mathrm{1}}−\frac{{x}\sqrt{{x}−\mathrm{1}}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}}{{x}+\mathrm{1}}= \\ $$$$… \\ $$$$=\frac{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)}} \\ $$
Answered by Ghisom last updated on 10/May/25
t=((√(x−1))/( (√(x+1)))) ⇔ x=−((t^2 +1)/(t^2 −1)) f(t)=−(((t^2 +1)t)/(t^2 −1))=(u/v) f ′(t)=((u′v−v′u)/v^2 )dt=−((t^4 −4t^2 −1)/((t^2 −1)^2 ))dt t=((√(x−1))/( (√(x+1)))) → dt=t′=(1/( (√((x+1)^3 (x−1))))) f ′(x)=((x^2 +x−1)/( (√((x+1)^3 (x−1)))))
$${t}=\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}}\:\Leftrightarrow\:{x}=−\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$${f}\left({t}\right)=−\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right){t}}{{t}^{\mathrm{2}} −\mathrm{1}}=\frac{{u}}{{v}} \\ $$$${f}\:'\left({t}\right)=\frac{{u}'{v}−{v}'{u}}{{v}^{\mathrm{2}} }{dt}=−\frac{{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} −\mathrm{1}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${t}=\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}}\:\rightarrow\:{dt}={t}'=\frac{\mathrm{1}}{\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)}} \\ $$$${f}\:'\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)}} \\ $$
Answered by Marzuk last updated on 10/May/25
1.(d/dx)[f(x).g(x)]=f ′(x)g(x)+f(x)g′(x) 2.(d/dx) [((f(x))/(g(x)))] = ((f ′(x)g(x)−f(x)g′(x))/([g(x)]^2 )) 3.(d/dx)[f(x)^n ] = nf(x)^(n−1) . f ′(x) ∴ (df/dx) = (√((x−1)/(x+1))) + x (d/dx)[(√((x−1)/(x+1)))] [app. 1.] =(√(((x−1)/(x+1)) )) + x[(1/2)((√((x−1)/(x+1))))^(−(1/2)) . (((x+1)−(x−1))/((x+1)^2 ))] [app. 2. 3.] ∴
$$\mathrm{1}.\frac{{d}}{{dx}}\left[{f}\left({x}\right).{g}\left({x}\right)\right]={f}\:'\left({x}\right){g}\left({x}\right)+{f}\left({x}\right){g}'\left({x}\right) \\ $$$$\mathrm{2}.\frac{{d}}{{dx}}\:\left[\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\right]\:=\:\frac{{f}\:'\left({x}\right){g}\left({x}\right)−{f}\left({x}\right){g}'\left({x}\right)}{\left[{g}\left({x}\right)\right]^{\mathrm{2}} } \\ $$$$\mathrm{3}.\frac{{d}}{{dx}}\left[{f}\left({x}\right)^{{n}} \right]\:=\:{nf}\left({x}\right)^{{n}−\mathrm{1}} \:.\:{f}\:'\left({x}\right) \\ $$$$\therefore\:\frac{{df}}{{dx}}\:=\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:+\:{x}\:\frac{{d}}{{dx}}\left[\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\right]\:\left[{app}.\:\mathrm{1}.\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\:}\:+\:{x}\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} .\:\frac{\left({x}+\mathrm{1}\right)−\left({x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right]\:\left[{app}.\:\mathrm{2}.\:\mathrm{3}.\right] \\ $$$$\therefore\: \\ $$

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