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y-x-2-1-ln-1-x-1-1-x-2-dy-dx-




Question Number 108237 by Study last updated on 15/Aug/20
y=(√(x^2 +1))−ln((1/x)+(√(1+(1/x^2 ))))  (dy/dx)=?
$${y}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{ln}\left(\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right) \\ $$$$\frac{{dy}}{{dx}}=? \\ $$
Answered by Dwaipayan Shikari last updated on 15/Aug/20
y=(√(x^2 +1))−log(1+(√(x^2 +1)))+logx  y=(x/( (√(x^2 +1))))−(x/( (√(x^2 +1))+x^2 +1))+(1/x)
$${y}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{log}\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)+{logx} \\ $$$${y}=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}} \\ $$
Answered by john santu last updated on 15/Aug/20
     ((♠JS♠)/(#•#))   y = (√(x^2 +1))−ln (((1+(√(1+x^2 )))/x))  (dy/dx) = (x/( (√(x^2 +1))))−((x/(1+(√(1+x^2 )))))((((x^2 /( (√(1+x^2 ))))−1−(√(1+x^2 )))/x^2 ))  =(x/( (√(x^2 +1))))−((x/(1+(√(1+x^2 )))))(((x^2 −(√(1+x^2 ))−1−x^2 )/(x^2 (√(1+x^2 )))))  =(x/( (√(x^2 +1))))+((1/(1+(√(1+x^2 )))))(((1+(√(1+x^2 )))/(x(√(1+x^2 )))))  =(x/( (√(x^2 +1))))+(1/(x(√(x^2 +1))))=((x^2 +1)/(x(√(x^2 +1))))  =((√(x^2 +1))/x)
$$\:\:\:\:\:\frac{\spadesuit{JS}\spadesuit}{#\bullet#} \\ $$$$\:{y}\:=\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}−\left(\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\left(\frac{\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }\right) \\ $$$$=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}−\left(\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\left(\frac{{x}^{\mathrm{2}} −\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}+\left(\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}} \\ $$
Commented by bemath last updated on 15/Aug/20
the simple answer
$${the}\:{simple}\:{answer} \\ $$
Answered by mathmax by abdo last updated on 15/Aug/20
y(x) =(√(x^2 +1))−ln((1/x)+(√(1+(1/x^2 )))) ⇒  y^′ (x) =(x/( (√(x^2 +1)))) +((((1/x)+(√(1+x^(−2) )))^′ )/((1/x)+(√(1+x^(−2) )))) =(x/( (√(x^2  +1)))) +((−(1/x^2 )+((−2x^(−3) )/(2(√(1+x^(−2) )))))/((1/x)+(√(1+x^(−2) ))))  =(x/( (√(1+x^2 ))))−(((1/x^2 )+(1/(x^3 (√(1+x^(−2) )))))/((1/x)+(√(1+x^(−2) )))) =(x/( (√(1+x^2 ))))−(1/x^3 ).((x+(1/( (√(1+x^(−2) )))))/(((1/x)+(√(1+x^(−2) )))))  =(x/( (√(1+x^2 ))))−((x(√(1+x^(−2) ))+1)/( (√(1+x^(−2) ))(x^2  +x^3 (√(1+x^2 )))))
$$\mathrm{y}\left(\mathrm{x}\right)\:=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\right)\:\Rightarrow \\ $$$$\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\:+\frac{\left(\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }\right)^{'} }{\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }}\:=\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}}\:+\frac{−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{−\mathrm{2x}^{−\mathrm{3}} }{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }}}{\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }} \\ $$$$=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}−\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }}}{\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }}\:=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }.\frac{\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }}}{\left(\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}−\frac{\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }+\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{2}} }\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}^{\mathrm{3}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)} \\ $$

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