分析:(1)先將式子展開化簡,再由冪函數(shù)的導(dǎo)數(shù)求解即可.
(2)由兩個函數(shù)商的求導(dǎo)法則,結(jié)合對數(shù)函數(shù)的導(dǎo)數(shù)糾結(jié)即可.
(3)將tanx轉(zhuǎn)化為正弦和余弦商的形式,由兩個函數(shù)商的求導(dǎo)法則求解.
(4)首先將函數(shù)看作兩個函數(shù)y=x和y=e1-cosx的乘積形式,利用兩個函數(shù)積的求導(dǎo)法則求解,
而y=e1-cosx為復(fù)合函數(shù),求導(dǎo)時應(yīng)用復(fù)合函數(shù)求導(dǎo)法則.
解答:解:(1)∵y=(1-
)(1+
)=
-
=
x--x,
∴y′=(x-
)′-(x
)′=-
x-
-
x-
.
(2)y′=(
)′=
=
•x-lnx,x2)=
.
(3)y′=(
)′=
(sinx)′cosx-sinx(cosx)′ |
cos2x |
=
cosxcosx-sinx(-sinx) |
cos2x |
=
.
(4)y′=(xe
1-cosx)′=e
1-cosx+x(e
1-cosx)′
=e
1-cosx+x[e
1-cosx•(1-cosx)′]
=e
1-cosx+xe
1-cosx•sinx
=(1+xsinx)e
1-cosx.
點評:本題考查導(dǎo)數(shù)的求解、導(dǎo)數(shù)的運算法則、復(fù)合函數(shù)的導(dǎo)數(shù),考查運算能力.