1、 1.2.2復(fù)合函數(shù)的求導(dǎo)法則教學(xué)目標(biāo)理解并掌握復(fù)合函數(shù)的求導(dǎo)法則.教學(xué)重點(diǎn)復(fù)合函數(shù)的求導(dǎo)方法：復(fù)合函數(shù)對(duì)自變量的導(dǎo)數(shù)，等于已知函數(shù)對(duì)中間變量的導(dǎo)數(shù)乘以中間變量對(duì)自變量的導(dǎo)數(shù)之積.教學(xué)難點(diǎn)正確分解復(fù)合函數(shù)的復(fù)合過程，做到不漏，不重，熟練，正確.一.創(chuàng)設(shè)情景(一)基本初等函數(shù)的導(dǎo)數(shù)公式表函數(shù)導(dǎo)數(shù)y =c-y = 0-n- *y = f (x) = x (n w Q )ny = nxy =sin xy = cosxy = cosxy = -sin xy = f (x) =axx .y = a ln a (a 0)一-一xy = f (x) = exy =ef (x) = log ax.1If (x

2、) = loga xf (x) =(a 0且a。1)x ln af (x) =ln x1 f (x)-x(二)導(dǎo)數(shù)的運(yùn)算法則導(dǎo)數(shù)運(yùn)算法則1 . f (x) 土g(x) =f (x) 土g (x)2. f (x) g(x) = f(x)g(x) 土 f(x)g(x)3.flxgfg。)(2)推論：Icf (x) 1 =cf (x)(常數(shù)與函數(shù)的積的導(dǎo)數(shù)，等于常數(shù)乘函數(shù)的導(dǎo)數(shù))二.新課講授復(fù)合函數(shù)的概念一般地，對(duì)于兩個(gè)函數(shù) y = f (u)和u = g(x),如果通過變量u , y可以表示成 x的函數(shù)，那么稱這個(gè)函數(shù)為函數(shù)y = f(u)和u = g(x)的復(fù)合函數(shù)，記作y = f (g(x)復(fù)

3、合函數(shù)的導(dǎo)數(shù)復(fù)合函數(shù)y = f (g(x)的導(dǎo)數(shù)和函數(shù) y = f (u)和u = g(x)的導(dǎo)數(shù)間的關(guān)系為V； = yu* ux即y對(duì)x的導(dǎo)數(shù)等于y Xu的導(dǎo)數(shù)與u對(duì)x的導(dǎo)數(shù)的乘積.若 y = f (g(x)，則 y= f (g(x) )1= f (g(x) )gx)三.典例分析例1求y =sin (tanx2)的導(dǎo)數(shù).【點(diǎn)評(píng)】求復(fù)合函數(shù)的導(dǎo)數(shù)，關(guān)鍵在于搞清楚復(fù)合函數(shù)的結(jié)構(gòu)，明確復(fù)合次數(shù)，由外層向內(nèi)層逐層求導(dǎo)，直到關(guān)于自變量求導(dǎo)，同時(shí)應(yīng)注意不能遺漏求導(dǎo)環(huán)節(jié)并及時(shí)化簡(jiǎn)計(jì)算結(jié)果.例2求y = xa 的導(dǎo)數(shù). .x2 -2ax【點(diǎn)評(píng)】本題練習(xí)商的導(dǎo)數(shù)和復(fù)合函數(shù)的導(dǎo)數(shù).求導(dǎo)數(shù)后要予以化簡(jiǎn)整理.【解法

4、一】y = sin 4x+ cos x= (sin x例3求y =sin4x + cos 4x的導(dǎo)數(shù).+ cos2x)2 2sin2cos2x= 1 sin22 x= 1-1 (1 - cos 4 x)42_ 3 , 1, 一十一cos 4 x. y = sin 4 x.44【解法二】v =(sin4x) +(cos4x) =4 sin 3x(sin x) +4 cos3x (cos x) =4 sin 3x cos x +4 cos3x ( sin x) =4 sin x cos x (sin 2x cos2x)【點(diǎn)評(píng)】解法一是先化簡(jiǎn)變形, 求導(dǎo)數(shù)，應(yīng)注意不漏步.=2 sin 2 x cos

5、 2 x= sin 4 x簡(jiǎn)化求導(dǎo)數(shù)運(yùn)算，要注意變形準(zhǔn)確.解法二是利用復(fù)合函數(shù)例4曲線y =x (x +1) (2 x)有兩條平行于直線 y =*的切線，求此二切線之間的距離.【解】y = x3+x2+2 x y = 3 x2+2 x+2令 y =1 即 3 x2 2 x 1=0,解得 x = 1 或 x = 1 .3114于是切點(diǎn)為P (1, 2), Q (，一二)，327過點(diǎn)P的切線方程為，y -2 = x 1即 x -y +1 = 0.顯然兩切線間的距離等于點(diǎn)Q到此切線的距離，故所求距離為,1 14一 1|3 27.216 227四.課堂練習(xí)1.求下列函數(shù)的導(dǎo)數(shù)(1) y =sinx3+sin33x; (2