Question

求导…………

Answer 1

如上图所示。

追答

Answer 2

(1)
y=(1+x^2).arctanx
y'
=(1+x^2)[1/(1+x^2)] + 2x.arctanx
=1 + 2x.arctanx
(2)
y=x.e^(x^2)

y' = (1 + 2x^2)e^(x^2)
(3)
e^y +xy =e^2
e^y. y' + xy' + y = 0
(e^y +x) y' = -y
y' = -y/(e^y +x)

追问

二阶求导

追答

这样
(1)
y=(1+x^2).arctanx
y'=(1+x^2)[1/(1+x^2)] + 2x.arctanx
=1 + 2x.arctanx
y''= 2x/(1+x^2) + 2arctanx
(2)
y=x.e^(x^2)
y' = (1 + 2x^2)e^(x^2)
y''= [4x + 2x(1+2x^2) ]e^(x^2)
= ( 4x^2+6x )e^(x^2)
(3)
e^y +xy =e^2
e^y. y' + xy' + y = 0
(e^y +x) y' = -y
y' = -y/(e^y +x)
y''
=[ y(e^y .y' + 1) -(e^y +x)y' ]/(e^y +x)^2
={ y ( e^y . [-y/(e^y +x)] + 1 ) -(e^y +x)[ -y/(e^y +x)] }/(e^y +x)^2
=[ y ( -ye^y +e^y +x ) +y(e^y +x) ] /(e^y +x)^3
= y ( - ye^y + 2e^y +2x ) /(e^y +x)^3

Answer 3

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