數(shù)學(xué)】334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)新人教A版選修課件

1、數(shù)學(xué)】334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)新人教A版選修數(shù)學(xué)】334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)新人教A版選修3.3.4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)數(shù)學(xué)】334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)新人教A版選修教學(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)用數(shù)學(xué)】334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)新人教A版選修由定義求導(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 xxf

2、xxfxy算比值.lim)3(0 xyyx 求求極極限限注意注意: :0)()(0 xxxfxf數(shù)學(xué)】334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)新人教A版選修常見函數(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)(cos公式：公式：還有必要建立求導(dǎo)法則，若兩個(gè)函數(shù)的導(dǎo)數(shù)存在，還有必要建立求導(dǎo)法則，若兩個(gè)函數(shù)的導(dǎo)數(shù)存在，如何求這兩個(gè)函數(shù)的和，差，積，商的導(dǎo)數(shù)呢？如何求這兩個(gè)函數(shù)的和，差，積，商的導(dǎo)數(shù)呢？數(shù)學(xué)】334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)新人教A版選修若若u=u(x)，v

3、=v(x)在在x處可導(dǎo)，則處可導(dǎo)，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(gè)函數(shù)的和兩個(gè)函數(shù)的和(或差或差)的導(dǎo)數(shù)的導(dǎo)數(shù),等于這兩個(gè)函數(shù)的導(dǎo)數(shù)等于這兩個(gè)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)四則運(yùn)算根據(jù)導(dǎo)數(shù)的定義，可以推出可導(dǎo)函數(shù)四則運(yùn)算的的求導(dǎo)法則求導(dǎo)法則數(shù)學(xué)】334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)新人教A版選修1.和和(或差或差)的導(dǎo)數(shù)的導(dǎo)數(shù)vuvu )( )()()(xvxuxfy證明： )()()()(xvxuxxvxxuy )()()()(xvxxvxuxxuvuxvxuxyxvxuxvxuxyxxxx0000liml

4、imlimlim)()(xvxu數(shù)學(xué)】334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)新人教A版選修的導(dǎo)數(shù)求例xxysin. 13的導(dǎo)數(shù)求例3. 224xxxy數(shù)學(xué)】334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)新人教A版選修2.積的導(dǎo)數(shù)積的導(dǎo)數(shù)法則法則2 兩個(gè)函數(shù)的積的導(dǎo)數(shù)兩個(gè)函數(shù)的積的導(dǎo)數(shù),等于第一個(gè)函數(shù)的導(dǎo)數(shù)乘第二等于第一個(gè)函數(shù)的導(dǎo)數(shù)乘第二個(gè)函數(shù)個(gè)函數(shù),加上第一個(gè)函數(shù)乘第二個(gè)函數(shù)的導(dǎo)數(shù)加上第一個(gè)函數(shù)乘第二個(gè)函數(shù)的導(dǎo)數(shù),即即vuvuvu )( )()()(xvxuxfy證明：)()()()(xvxuxxvxxuy)()()()()()()()(xvxuxxvxuxxvxuxxvxxuxx

5、vxxvxuxxvxxuxxuxy)()()()()()(從而時(shí)，于是當(dāng)處連續(xù)，處可導(dǎo)，所以它在點(diǎn)在點(diǎn)因?yàn)?.()(0)(xvxxvxxxxv數(shù)學(xué)】334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)新人教A版選修xxvxxvxuxxvxxuxxuxyxxx)()(lim)()()()(limlim000)()()()(xvxuxvxu)(uvvuuvy即uCCu )( :推論數(shù)學(xué)】334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)新人教A版選修的導(dǎo)數(shù)求例4532. 322xxxy的導(dǎo)數(shù)求例)23)(32(. 42xxy數(shù)學(xué)】334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)新人教A版選修3.商的導(dǎo)數(shù)商的導(dǎo)數(shù)法則法則3 兩個(gè)函數(shù)的商的導(dǎo)數(shù)兩個(gè)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數(shù)學(xué)】334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)新人教A版選修處的導(dǎo)數(shù)在點(diǎn)求例333. 62