高中數(shù)學(xué)第二章 4_2 導(dǎo)數(shù)的乘法與除法法則 課件

第二章 變化率與導(dǎo)數(shù)4.2 導(dǎo)數(shù)的乘法與除法法則求函數(shù)的導(dǎo)數(shù)的步驟是怎樣的?(1)()();yf xxf x 求函數(shù)的增量(2):()();yf xxf xxx求函數(shù)的增量與自變量的增量的比值0(3)()lim.xyyfxx 求極限，得導(dǎo)函數(shù)導(dǎo)數(shù)公式表(其中三角函數(shù)的自變量單位是弧度).()()()(),()()()(xgxfxgxfxgxfxgxf兩個(gè)函數(shù)和(差)的導(dǎo)數(shù)等于這兩個(gè)函數(shù)導(dǎo)數(shù)的和(差),即:導(dǎo)數(shù)的加法和減法法則是什么?問(wèn)題提出?),()()(2如何來(lái)求它的導(dǎo)數(shù)呢如果有函數(shù)xfxxgxfy分析推導(dǎo)),()()(,:0200200 xfxxxfxxyxx量可以得到函數(shù)值的改變的一個(gè)改變量首先給定自變量按照求函數(shù)導(dǎo)數(shù)的步驟),()()()()()()()()()()()()(020200020020200020020020 xfxxxxxxfxxfxxxxfxxxxfxxfxxxxfxxxfxxxy寫(xiě)成相應(yīng)的平均變化率可以).()()()(,).(2)()()()(,2)(lim),()()(lim,)(lim,02220002002020200000020200 xfxxfxxfxxfxxfxxxfxxgxfxxxxxxfxxfxxfxxxxxxx的導(dǎo)數(shù)為因此處的導(dǎo)數(shù)值為在知由于令:),()()()(,我們有和是的導(dǎo)數(shù)分別和若兩個(gè)函數(shù)一般地xgxfxgxf.)()()()()()()(),()()()()()(2xgxgxfxgxfxgxfxgxfxgxfxgxf).()(:,)(,xf kxkfkxg有時(shí)當(dāng)特別地.)()()()(),()()()(xgxfxgxfxgxfxgxf例題講解2.:(1);(2)sin1;(3)ln.xyx eyxxyxx求下面函例數(shù)的導(dǎo)數(shù).)2(2)(:,)(,2)(,)()()1(:22222xxxxxxxexxexxeexexgxxfexgxxfexy可得法則根據(jù)兩函數(shù)之積的求導(dǎo)由導(dǎo)數(shù)公式表分別得出之積與是函數(shù)函數(shù)解.cos2sin)sin(:,cos)(,21)(,sin)()(sin)2(xxxxxxxxgxxfxxgxxfxxy可得法則根據(jù)兩函數(shù)之積的求導(dǎo)由導(dǎo)數(shù)公式表分別得出之積與是函數(shù)函數(shù).1ln1ln1)ln(:,1)(,1)(,ln)()(ln)3(xxxxxxxxgxfxxgxxfxxy可得法則根據(jù)兩函數(shù)之積的求導(dǎo)由導(dǎo)數(shù)公式表分別得出之積與是函數(shù)函數(shù)2.:sin(1),(2)2.lnxxyyxx求下列函數(shù)的導(dǎo)數(shù)例.sincos1sincossin:,1)(,cos)(:,)(sin)(sin)1(:22xxxxxxxxxxxgxxfxxgxxfxxy由求導(dǎo)的除法法則得出根據(jù)導(dǎo)數(shù)公式表分別得之商和函數(shù)是函數(shù)函數(shù)解例題講解.ln)1ln2()(ln1ln2ln:,1)(,2)(:,ln)()(ln)2(222222xxxxxxxxxxxxgxxfxxgxxfxxy由求導(dǎo)的除法法則得出根據(jù)導(dǎo)數(shù)公式表分別得之商和函數(shù)是函數(shù)函數(shù)22.:cos(13)(lnsin);(2).xxyxxxyx求下列函數(shù)的導(dǎo)數(shù)例,cos1)(,2)(:.sinln)()()sin(ln)1(:22xxxgxxfxxxgxxfxxxy導(dǎo)法則可得的求由導(dǎo)數(shù)公式表及和函數(shù)的積與是函數(shù)函數(shù)解.cossin2ln2cos1)sin(ln2)sin(ln:222xxxxxxxxxxxxxxxx由求導(dǎo)的乘法法則可得例題講解,2)(,1sin)(:.)(cos)(cos)2(22xxgxxfxxgxxxfxxxy導(dǎo)法則可得的求由導(dǎo)數(shù)公式表及差函數(shù)的商與是函數(shù)函數(shù).cos2sin2cos2sin1)(2)(cos1sincos:332222xxxxxxxxxxxxxxxxxxx用求導(dǎo)的除法法則可得練習(xí).)0,1(ln2)(的切線方程過(guò)點(diǎn)求曲線xxxfx.2ln2ln212ln2ln21ln2ln2)(:.ln2)(:xxxxxxxxfxxxfxxxxxxx的四則運(yùn)算法則可得根據(jù)導(dǎo)數(shù)公式表和導(dǎo)數(shù)的導(dǎo)函數(shù)首先求出函數(shù)解).1(3)0,1(ln2)(.3121ln2ln21)1(),(111xyxxxffxfxx的切線方程為過(guò)點(diǎn)曲線得所求切線斜率代入將導(dǎo)數(shù)的乘法與除法法則是什么?.)()()()()()()(),()()()()()(2xgxgxfxgxfxgxfxgxfxgxfxgxf