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Question Number 166829 by mnjuly1970 last updated on 01/Mar/22
calculate If , f(x)=(( (x^2 +1)(((x^( 2) +x−2)(x^( 4) −1)(x^( 2) +2x−3)+16))^(1/3) + (√(x^( 2) +3)))/(( 1+x +x^( 2) ))) then , f ′ (1 ) =? ■ m.n
$$ \\ $$$$\:\:\:\:\:\:\:\:{calculate}\: \\ $$$$\:\:\:\mathrm{I}{f}\:,\:\:\:\:{f}\left({x}\right)=\frac{\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt[{\mathrm{3}}]{\left({x}^{\:\mathrm{2}} +{x}−\mathrm{2}\right)\left({x}^{\:\mathrm{4}} −\mathrm{1}\right)\left({x}^{\:\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}\right)+\mathrm{16}}\:\:+\:\sqrt{{x}^{\:\mathrm{2}} +\mathrm{3}}}{\left(\:\mathrm{1}+{x}\:+{x}^{\:\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{then}\:,\:\:\:\:{f}\:'\:\left(\mathrm{1}\:\right)\:=?\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$ \\ $$
Answered by mathsmine last updated on 01/Mar/22
f(x)=(((x^2 +1)((g(x)))^(1/3) +(√(x^2 +3)))/((1+x+x^2 ))) f′(x)=(((2x((g(x)))^(1/3) +((g′(x))/(3g^(2/3) (x)))(x^2 +1)+(x/( (√(x^2 +3))))))/(1+x+x^2 ))−(((1+2x)((x^2 +1)((g(x)))^(1/3) +(√(x^2 +3))))/((1+x+x^2 )^2 )) f′(1)=((2((16))^(1/3) +2((g′(1))/(3(16)^(2/9) ))+(1/2))/3)−((2((16))^(1/3) +2)/3) g(x)=(x^2 +x−2)(x^4 −1)(x^2 +2x−3)+16 (fgh)^′ =f′gh+fg′h+fgh′ 1 root of x^2 +x−2,x^4 −1,x^2 +2x−3⇒ g′(1)=0 f′(1)=((2((16))^(1/3) +(1/2))/3)−((2((16))^(1/3) +2)/3)=−(1/2)
$${f}\left({x}\right)=\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt[{\mathrm{3}}]{{g}\left({x}\right)}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)} \\ $$$${f}'\left({x}\right)=\frac{\left(\mathrm{2}{x}\sqrt[{\mathrm{3}}]{{g}\left({x}\right)}+\frac{{g}'\left({x}\right)}{\mathrm{3}{g}^{\frac{\mathrm{2}}{\mathrm{3}}} \left({x}\right)}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}\right)}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }−\frac{\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt[{\mathrm{3}}]{{g}\left({x}\right)}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${f}'\left(\mathrm{1}\right)=\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{16}}+\mathrm{2}\frac{{g}'\left(\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{16}\right)^{\frac{\mathrm{2}}{\mathrm{9}}} }+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{16}}+\mathrm{2}}{\mathrm{3}} \\ $$$${g}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}−\mathrm{2}\right)\left({x}^{\mathrm{4}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}\right)+\mathrm{16} \\ $$$$\left({fgh}\right)^{'} ={f}'{gh}+{fg}'{h}+{fgh}' \\ $$$$\mathrm{1}\:{root}\:{of}\:{x}^{\mathrm{2}} +{x}−\mathrm{2},{x}^{\mathrm{4}} −\mathrm{1},{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}\Rightarrow \\ $$$${g}'\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}'\left(\mathrm{1}\right)=\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{16}}+\mathrm{2}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

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