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2.3.1 复合函数的求导法则
定理(链式法则) 若函数u=φ(x)在点x处可导,函数y=f(u)在其对应点u=φ(x)处也可导,则复合函数y=f(φ(x))在点x处可导,且
![](https://epubservercos.yuewen.com/6E8314/14615890804593006/epubprivate/OEBPS/Images/img00060002.jpg?sign=1738888230-RdB9E0nsHNm0gDUh6EGBMUmacpd7nz4Q-0-1439cd8cdf806ac899b568c3a170a0c0)
简记为
y′x=y′u·u′x.
上述公式称为复合函数的求导法则,也称链式法则.
证明 设x,u,y的改变量分别为Δx,Δu,Δy,因为函数u=φ(x)在点x处可导,所以u=φ(x)在点x处连续,即当Δx→0时,Δu→0,且假设Δu≠0时:
![](https://epubservercos.yuewen.com/6E8314/14615890804593006/epubprivate/OEBPS/Images/img00060003.jpg?sign=1738888230-193HvlQtjYVzErEQHJcVMKkpD5DKhpg1-0-9f57a90d0c219d103026c2c464de1be2)
![](https://epubservercos.yuewen.com/6E8314/14615890804593006/epubprivate/OEBPS/Images/img00060004.jpg?sign=1738888230-IRk9tQh62U9z4D5bcEQpykPNgAvZksBV-0-d385d1188047dcd37c1fd4bb11b38360)
即
[f(φ(x))]′=f′(u)·φ′(x).
例1 设y=(2x+1)5,求y′.
解 该函数由y=u5,u=2x+1复合而成,所以
y′x=y′u·u′x=(u5)′u·(2x+1)′x=(5u4)·2=10(2x+1)4.
例2 设y=ln(1+x2),求y′.
解 该函数由y=lnu,u=1+x2复合而成,所以
![](https://epubservercos.yuewen.com/6E8314/14615890804593006/epubprivate/OEBPS/Images/img00060005.jpg?sign=1738888230-tnhr8nrCgJ5qyovB8I0tSaXgguNHi42i-0-df9d9bcab8b91ebe54bf8d18ae02c949)
发现:今后在求复合函数的导数时,要达到熟练运用该求导法则,应熟练运用如下方法技巧。“将复合过程默记在心、不必写出、由外往里、逐层求导”.
例如,求复合函数y=sin(3x2-5)的导数,将函数的复合过程y=sinu,u=3x2-5默记在心、不必写出,由外往里、逐层求导,得
y′=[sin(3x2-5])′=cos(3x2-5)·(3x2-5)′=6xcos(3x2-5).
例3 求下列函数的导数y′.
(1);(2)
;
(3)y=sinln(1-2x);(4)y=sec32x.
解
![](https://epubservercos.yuewen.com/6E8314/14615890804593006/epubprivate/OEBPS/Images/img00061001.jpg?sign=1738888230-72jaIqrtCpl0aas1gd2u4iqCRjfmuzQZ-0-98a19b458145ec095055eab3dca41697)
(4)y′=(sec32x)′=3sec22x·(sec2x)′
=3sec22x·sec2x·tan2x·(2x′)=6sec32x·tan2x.
发现:如果y=f(u),u=φ(v),v=ψ(x)均可导,则复合函数y=f(φ(ψ(x)))可导,且y)′x=y′u·u′v·v′x.
此求导法则可以推广到有限个函数复合而成的复合函数求导.
例4 ,求y′
![](https://epubservercos.yuewen.com/6E8314/14615890804593006/epubprivate/OEBPS/Images/img00061006.jpg?sign=1738888230-MZh59kPE4ZcVnxPVYaBcWegPFEWXkMTD-0-c1da9debbc62d4f17ab601906be35ba7)