3.1導(dǎo)數(shù)導(dǎo)學(xué)案

1、導(dǎo)數(shù)的概念及運(yùn)算一、預(yù)習(xí)案(-)高考解讀通過(guò)圖像能利用給出的基本初等函數(shù)的導(dǎo)數(shù)公式求簡(jiǎn)單函數(shù)的導(dǎo)數(shù), 直觀地理解導(dǎo)數(shù)的幾何意義，會(huì)求在某點(diǎn)和過(guò)某點(diǎn)的切線方程。(-)知識(shí)清單1、基本初等函數(shù)的導(dǎo)數(shù)公式原函數(shù)導(dǎo)函數(shù)f (x) = xn(n 花 Q)f (x) =sin xf (x) =cosxf (x) = axf (x) = exf (x)=logaXf (x) = In x2、求導(dǎo)法則運(yùn)算(1)f(x)-g(x) (2)f(x) g(x)=(3)他二_g(x)復(fù)合函數(shù)的導(dǎo)數(shù)：設(shè)u=v(x)在x處可導(dǎo)，y二f(u)在點(diǎn)u處可導(dǎo),則復(fù)合函數(shù)fv(x)在點(diǎn)x處可導(dǎo)，且f (x)=(三)預(yù)期效果及存在

2、困惑二、導(dǎo)學(xué)案(-)完成新亮劍(紅色)第50頁(yè)查缺補(bǔ)漏。(-)高考類型考點(diǎn)一、導(dǎo)數(shù)運(yùn)算，兀仁 已知函數(shù)f (x)- xsinxax,且f ()=1,則a的值等于()2A.OB. 1C.2D.42、函數(shù)f(x)的定義域是R, f (0) = 2,對(duì)任意x R, f (x) f (x)1則不等式ex f(x) ex, 1的解集為考點(diǎn)二、導(dǎo)數(shù)幾何意義的應(yīng)用3、已知函數(shù) f (x) = x3 -4x2 5x -4o(1) 求曲線f (x)在點(diǎn)(2, f (2)處的切線方程；(2) 求經(jīng)過(guò)點(diǎn)A(2,2)的曲線f(x)的切線方程。練習(xí)：321 (2018課標(biāo)I)設(shè)函數(shù)f(x)二x, (a-1)x aXo若

3、f(x)為奇函數(shù),A. y 2xb. y = _xc. y = 2x D. y = x2. (2017威海質(zhì)檢)已知函數(shù)f(x)二xlnx,若直線I過(guò)點(diǎn)(0, 1),并且與 曲線y二f(x)相切，則直線I的方程為()A.x+ y 1 = 0 Bx y 1 二 0Cx+y+1二0Dx一y+1=0課堂總結(jié):三、鞏固案1.(2016北京節(jié)選)設(shè)函數(shù)f(x) - xeAFx,曲線y=f(x)在(2, f)處的切線方程為y = (e)x 4,求a, b的值。2. (2015全國(guó)11)設(shè)函數(shù)f(X)是奇函數(shù)f(X)的導(dǎo)函數(shù)，f(1)=0,當(dāng)x 0 時(shí)，xf (x) - f (x) : :0,解不等式 f (x)0。3 (2016課標(biāo)II)若直線y二kxb是曲線y=1nx2的切線，也是曲線y =ln(x 1)的切線，貝 u b=4. 已知函數(shù) f(x)=lnx 與 g(x)二 kx b(k,b R)。d )求(刈在x