设f(x)=g(x)?e?xx,x≠00,x=0其中g(x)有二阶连续导数,且g(0)=1,g′(0)=-1.(1)求f′(x)

2025-05-16 11:51:10
推荐回答(1个)
回答1:


(1)
当x≠0时,
f′(x)=

x[g′(x)+e?x]?g(x)+e?x
x2
xg′(x)?g(x)+(x+1)e?x
x2

当x=0时,由导数定义,有:
f′(0)=
lim
x→0
f(x)?f(0)
x?0
lim
x→0
g(x)?e?x
x2
=
lim
x→0
g′(x)+e?x
2x
=
lim
x→0
g″(x)?e?x
2
g″(0)?1
2

所以:
f′(x)=
xg′(x)?g(x)+(x+1)e?x
x2
x≠0
g″(0)?1
2
x=0


(2)
因为在x=0处,有:
lim
x→0
f′(x)=
lim
x→0
xg′(x)?g(x)+(x+1)e?x
x2

=
lim
x→0
g′(x)+xg″(x)?g′(x)+e?x?(x+1)e?x
2x

=
lim
x→0
g″(x)?e?x
2