3.求下列函数的二阶导数.-|||-(1) =ln (1+(x)^2)-|||-; . (2) =sin 2xcdot (e)^x ;-|||-(3) =xcos x; ...-|||-(4) =dfrac (1)(1-{x)^2} ;-|||-(5) =(x)^2(e)^x ;-|||-(6) =cos xln x ;-|||-(7) =xln x ;-|||-(8) =dfrac ({e)^x}(x) ,-|||-(9) =(e)^-xcdot sin x ; .-|||-(10) =(e)^2x-1 -

$\color{red}{（}1\color{red}{）}{y}^{''}=\dfrac{2-2{x}^{2}}{\color{red}{（}{x}^{2}+1\color{red}{）}{}^{2}}$

$\color{red}{（}2\color{red}{）}{y}^{''}={e}^{x}\color{red}{（}4\cos 2x-3\sin 2x\color{red}{）}$

$\color{red}{（}\color{red}{3}）{y}^{''}\color{red}{=}\color{red}{-}\color{red}{2}\color{red}{\sin }\color{red}{x}\color{red}{-}\color{red}{x}\color{red}{\cos }\color{red}{x}$

$\left(4\right)y''=\dfrac{4{x}^{2}-6{x}^{4}+2}{{\left(1-{x}^{2}\right)}^{4}}$

$\left(5\right)y''=\left({x}^{2}+4x+2\right){e}^{x}$

$\left(6\right)y''=-\cos x\ln \,x-\dfrac{2\sin x}{x}-\dfrac{\cos x}{{x}^{2}}$

$\left(7\right)y''=\dfrac{1}{x}$

$\left(8\right)y''=\dfrac{\left({x}^{2}-2x+2\right){e}^{x}}{{x}^{3}}$

$\left(9\right)y''={-2e}^{-x}\cos x$

$\left(10\right)y''=4{e}^{2x-1}$