高二數(shù)學(xué)選修幾常個(gè)用函數(shù)的導(dǎo)數(shù)及基本初等函數(shù)的導(dǎo)數(shù)公式教學(xué)課件

Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.1．知識(shí)與技能了解常數(shù)函數(shù)和冪函數(shù)的求導(dǎo)方法和規(guī)律，會(huì)求任意y＝xα(α∈Q)的導(dǎo)數(shù)．2．過(guò)程與方法掌握基本初等函數(shù)的導(dǎo)數(shù)公式，并能利用這些公式求基本初等函數(shù)的導(dǎo)數(shù)．

Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.本節(jié)重點(diǎn)：常數(shù)函數(shù)、冪函數(shù)的導(dǎo)數(shù)本節(jié)難點(diǎn)：由常見(jiàn)冪函數(shù)的求導(dǎo)公式發(fā)現(xiàn)規(guī)律，得到冪函數(shù)的求導(dǎo)公式．利用常見(jiàn)函數(shù)的導(dǎo)數(shù)公式可以比較簡(jiǎn)捷的求出函數(shù)的導(dǎo)數(shù)，其關(guān)鍵是牢記和運(yùn)用好導(dǎo)數(shù)公式．解題時(shí)認(rèn)真觀察函數(shù)的結(jié)構(gòu)特征，積極地進(jìn)行聯(lián)想化歸，才能抓住問(wèn)題的本質(zhì)，把解題思路放開(kāi)．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.1．在應(yīng)用(sinx)′＝cosx與(cosx)′＝－sinx時(shí)，一要注意函數(shù)的變化；二要注意符號(hào)的變化．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.1．若f(x)＝c，則f′(x)＝

.若f(x)＝xn(n∈N*)，則f′(x)＝

.2．若f(x)＝sinx，則f′(x)＝

.若f(x)＝cosx，則f′(x)＝

.3．若f(x)＝ax，則f′(x)＝ ．若f(x)＝ex，則f′(x)＝

.0nxn－1cosx－sinxaxlna(a>0)exEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.

[例1]求下列函數(shù)的導(dǎo)數(shù)．(1)y＝a2(a為常數(shù))．(2)y＝x12.(3)y＝cosx.[解析]

(1)∵a為常數(shù)，∴a2為常數(shù)，∴y′＝(a2)′＝0.(2)y′＝(x12)′＝12x11(3)y′＝(cosx)′＝－sinx.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.[點(diǎn)評(píng)](1)用導(dǎo)數(shù)的定義求導(dǎo)是求導(dǎo)數(shù)的基本方法，但運(yùn)算較繁．利用常用函數(shù)的導(dǎo)數(shù)公式，可以簡(jiǎn)化求導(dǎo)過(guò)程，降低運(yùn)算難度．(2)利用導(dǎo)數(shù)公式求導(dǎo)，應(yīng)根據(jù)所給問(wèn)題的特征，恰當(dāng)?shù)剡x擇求導(dǎo)公式，將題中函數(shù)的結(jié)構(gòu)進(jìn)行調(diào)整．如將根式、分式轉(zhuǎn)化為指數(shù)式，利用冪函數(shù)的求導(dǎo)公式求導(dǎo)．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.[點(diǎn)評(píng)]求函數(shù)在某點(diǎn)處的導(dǎo)數(shù)的步驟是先求導(dǎo)函數(shù)，再代入變量的值求導(dǎo)數(shù)．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.[點(diǎn)評(píng)]在確定與切線垂直的直線方程時(shí)，應(yīng)注意考察函數(shù)在切點(diǎn)處的導(dǎo)數(shù)y′是否為零，當(dāng)y′＝0時(shí)，切線平行于x軸，過(guò)切點(diǎn)P垂直于切線的直線斜率不存在．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.求曲線y＝3x2的斜率等于12的切線方程．[解析]設(shè)切點(diǎn)為P(x0，y0)，則y′＝(3x2)′＝6x，∴y′|x＝x0＝12，即6x0＝12，∴x0＝2當(dāng)x0＝2時(shí)，y0＝12∴切點(diǎn)P的坐標(biāo)為(2,12)∴所求切線方程為：y－12＝12(x－2)，即y＝12x－12.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.一、選擇題1．函數(shù)f(x)＝0的導(dǎo)數(shù)是 ()A．0 B．1C．不存在 D．不確定[答案]A[解析]

常數(shù)函數(shù)的導(dǎo)數(shù)為0Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.A．x－y－1＝0 B．x＋y－3＝0C．x－y＋1＝0 D．x＋y－1＝0[答案]

AEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.[答案]DEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.[答案]BEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.二、填空題5．曲線y＝xn在x＝2處的導(dǎo)數(shù)為12，則n等于________．[答案]3[解析]

y′＝nxn－1，∴y′|x＝2＝n2n－1＝12，∴n＝3.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.6．若函數(shù)y＝sint，則y′|t＝6π＝________.[答案]

1[解析]

y′＝(sint)′＝cost，y′|t＝6π＝cos6π＝1.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2019-2019AsposePtyLtd.三、解答題7．求拋物線y＝x2上的點(diǎn)到直線x－y－2＝0的最短距離．[解析]

平移直線x－y－2＝0與拋物線