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Question Number 38722 by maxmathsup by imad last updated on 28/Jun/18
let f(x)= (x+1)e^(−x) and g(x)=ln(2+x^2 ) 1) calculate fog(x) and gof(x) 2) calculate (fog)^′ (x) and (gof)^′ (x).
$${let}\:{f}\left({x}\right)=\:\left({x}+\mathrm{1}\right){e}^{−{x}} \:\:{and}\:\:{g}\left({x}\right)={ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{fog}\left({x}\right)\:{and}\:{gof}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\left({fog}\right)^{'} \left({x}\right)\:{and}\:\left({gof}\right)^{'} \left({x}\right). \\ $$
Commented by math khazana by abdo last updated on 30/Jun/18
1) fog(x)=f(g(x))=(g(x)+1)e^(−g(x)) ={ln(2+x^2 )+1} e^(−ln(2+x^2 )) = ((1+ln(2+x^2 ))/(2+x^2 )) gof(x)=g(f(x))= ln(2+f^2 (x)) =ln(2+(x+1)^2 e^(−2x) ) 2)(fog)^′ (x)=((((2x)/(2+x^2 ))(2+x^2 ) −2xln(2+x^2 ))/((2+x^2 )^2 )) = ((2x −2xln(2+x^2 ))/((2+x^2 )^2 )) . (gof)^′ (x) = ((2(x+1)e^(−2x) −2(x+1)^2 e^(−2x) )/(2+(x+1)^2 e^(−2x) )) =(({2x+2−2x^2 −4x−2}e^(−2x) )/(2+(x+1)^2 e^(−2x) )) =(((−2x^2 −2x)e^(−2x) )/(2+(x+1)^2 e^(−2x) )) = ((−2x^2 −2x)/(2 e^(2x) +(x+1)^2 )) .
$$\left.\mathrm{1}\right)\:{fog}\left({x}\right)={f}\left({g}\left({x}\right)\right)=\left({g}\left({x}\right)+\mathrm{1}\right){e}^{−{g}\left({x}\right)} \\ $$$$=\left\{{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)+\mathrm{1}\right\}\:{e}^{−{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)} =\:\frac{\mathrm{1}+{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\mathrm{2}+{x}^{\mathrm{2}} } \\ $$$${gof}\left({x}\right)={g}\left({f}\left({x}\right)\right)=\:{ln}\left(\mathrm{2}+{f}^{\mathrm{2}} \left({x}\right)\right) \\ $$$$={ln}\left(\mathrm{2}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} {e}^{−\mathrm{2}{x}} \right) \\ $$$$\left.\mathrm{2}\right)\left({fog}\right)^{'} \left({x}\right)=\frac{\frac{\mathrm{2}{x}}{\mathrm{2}+{x}^{\mathrm{2}} }\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}{xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{2}{x}\:−\mathrm{2}{xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:. \\ $$$$\left({gof}\right)^{'} \left({x}\right)\:=\:\frac{\mathrm{2}\left({x}+\mathrm{1}\right){e}^{−\mathrm{2}{x}} \:−\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} {e}^{−\mathrm{2}{x}} }{\mathrm{2}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:{e}^{−\mathrm{2}{x}} } \\ $$$$=\frac{\left\{\mathrm{2}{x}+\mathrm{2}−\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{4}{x}−\mathrm{2}\right\}{e}^{−\mathrm{2}{x}} }{\mathrm{2}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} {e}^{−\mathrm{2}{x}} } \\ $$$$=\frac{\left(−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right){e}^{−\mathrm{2}{x}} }{\mathrm{2}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} {e}^{−\mathrm{2}{x}} }\:\:=\:\frac{−\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{2}{x}}{\mathrm{2}\:{e}^{\mathrm{2}{x}} \:+\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:. \\ $$

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