为自己加油

利用一阶微分形式不变性及微分运算法则，求微分．
(1)y=(arctanx)³
(2)y=lnx/cos3x
(3)y=e^sin(1/x2)
(4)y=e^x2sin3x

(1)dy=3(arctanx)²d(arctanx) =3(arctan)²1/(1+x²)≥dx (2)dy=pd(lnx)•cos3x-lnxd(cos3x)]/cos²3x =[(1/x)dx•cos³x-lnx(-3sin3xdx)]/cos²3x ={[(1/x)cos3x+3sin3x•lnx]/cos²3x}dx (3)dy=e^sin(1/x2)d(sin1/x²) =e^sin(1/n2)•cos1/x²d(1/x²) =e^sin(1/x2)•cos1/x²(-2/x³)dx (4)dy=d(e^x2)•sin3x+e^x2d(sin3x) =e^x2•d(x²)•sin3x+e^x2•cos3x•d(3x) =e^x2•2xdx•sin3x+e^x2•cos3x•3dx =(2xe^x2sin3x+3e^x2cos3x)dx