《福建省長泰一中高中數(shù)學(xué) 3.3.3《導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用函數(shù)的和差積商的導(dǎo)數(shù)》課件 新人教A版選修11》由會員分享,可在線閱讀,更多相關(guān)《福建省長泰一中高中數(shù)學(xué) 3.3.3《導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用函數(shù)的和差積商的導(dǎo)數(shù)》課件 新人教A版選修11(15頁珍藏版)》請?jiān)谘b配圖網(wǎng)上搜索。
1、新人教版選修1-1全套課件3.3.導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用-函數(shù)的和差積商的導(dǎo)數(shù)教學(xué)教學(xué) 目標(biāo)目標(biāo)熟練運(yùn)用導(dǎo)數(shù)的函數(shù)的和差積商運(yùn)算法則,并能靈活運(yùn)用教學(xué)重點(diǎn)教學(xué)重點(diǎn):熟練運(yùn)用導(dǎo)數(shù)的四則運(yùn)算法則教學(xué)難點(diǎn)教學(xué)難點(diǎn):商的導(dǎo)數(shù)的運(yùn)用由定義求導(dǎo)數(shù)(三步法由定義求導(dǎo)數(shù)(三步法)步驟步驟:);()() 1 (00 xfxxfy求增量;)()()2(00 xxfxxfxy算比值.lim)3(0 xyyx 求極限求極限注意注意: :0)()(0 xxxfxf常見函數(shù)的導(dǎo)數(shù)公式:常見函數(shù)的導(dǎo)數(shù)公式:公式:公式:)(0為為常常數(shù)數(shù)CC )()(1Qnnxxnn公式:公式:xxcos)(sin公式:公式:xxsin)(
2、cos公式:公式:還有必要建立求導(dǎo)法則,若兩個函數(shù)的導(dǎo)數(shù)存在,還有必要建立求導(dǎo)法則,若兩個函數(shù)的導(dǎo)數(shù)存在,如何求這兩個函數(shù)的和,差,積,商的導(dǎo)數(shù)呢?如何求這兩個函數(shù)的和,差,積,商的導(dǎo)數(shù)呢?若若u=u(x),v=v(x)在在x處可導(dǎo),處可導(dǎo),則則vuvu )( 1.和和(或差或差)的導(dǎo)數(shù)的導(dǎo)數(shù)法則法則1 兩個函數(shù)的和兩個函數(shù)的和(或差或差)的導(dǎo)數(shù)的導(dǎo)數(shù),等于這兩個函數(shù)的導(dǎo)數(shù)等于這兩個函數(shù)的導(dǎo)數(shù)的和的和(或差或差),即即 根據(jù)導(dǎo)數(shù)的定義,可以推出可導(dǎo)函數(shù)根據(jù)導(dǎo)數(shù)的定義,可以推出可導(dǎo)函數(shù)四則運(yùn)算的四則運(yùn)算的求導(dǎo)法則求導(dǎo)法則1.和和(或差或差)的導(dǎo)數(shù)的導(dǎo)數(shù)vuvu )( )()()(xvxuxfy證
3、明: )()()()(xvxuxxvxxuy )()()()(xvxxvxuxxuvuxvxuxyxvxuxvxuxyxxxx0000limlimlimlim)()(xvxu的導(dǎo)數(shù)求例xxysin.13的導(dǎo)數(shù)求例3. 224xxxy2.積的導(dǎo)數(shù)積的導(dǎo)數(shù)法則法則2 兩個函數(shù)的積的導(dǎo)數(shù)兩個函數(shù)的積的導(dǎo)數(shù),等于第一個函數(shù)的導(dǎo)數(shù)乘第二等于第一個函數(shù)的導(dǎo)數(shù)乘第二個函數(shù)個函數(shù),加上第一個函數(shù)乘第二個函數(shù)的導(dǎo)數(shù)加上第一個函數(shù)乘第二個函數(shù)的導(dǎo)數(shù),即即vuvuvu )( )()()(xvxuxfy證明:)()()()(xvxuxxvxxuy)()()()()()()()(xvxuxxvxuxxvxuxxvxxu
4、xxvxxvxuxxvxxuxxuxy)()()()()()(從而時,于是當(dāng)處連續(xù),處可導(dǎo),所以它在點(diǎn)在點(diǎn)因?yàn)?.()(0)(xvxxvxxxxvxxvxxvxuxxvxxuxxuxyxxx)()(lim)()()()(limlim000)()()()(xvxuxvxu)(uvvuuvy即uCCu )( :推論的導(dǎo)數(shù)求例4532. 322xxxy的導(dǎo)數(shù)求例)23)(32(. 42xxy3.商的導(dǎo)數(shù)商的導(dǎo)數(shù)法則法則3 兩個函數(shù)的商的導(dǎo)數(shù)兩個函數(shù)的商的導(dǎo)數(shù),等于分子的導(dǎo)數(shù)與分母的積等于分子的導(dǎo)數(shù)與分母的積,減減去分母的導(dǎo)數(shù)與分子的積去分母的導(dǎo)數(shù)與分子的積,再除以分母的平方再除以分母的平方,即即)0()(2vvvuvuvu的導(dǎo)數(shù)例xxysin. 52xxxxxy222sin)(sinsin)(解:xxxxx22sincossin2處的導(dǎo)數(shù)在點(diǎn)求例333. 62xxxy222)3(2)3()3(1xxxxy解:222)3(36xxx6114424)39(3189|23xy