1、2.32.3反函數(shù)的導數(shù)，復合函數(shù)的求導法則反函數(shù)的導數(shù)，復合函數(shù)的求導法則一、反函數(shù)的導數(shù)一、反函數(shù)的導數(shù)設)(yx??是直接函數(shù)，)(xfy?是它的反函數(shù)，假定)(yx??在Iy內單調、可導，而且0)(??y?，則反函數(shù))(xfy?在間)(|yxIyyxxI????內也是單調、可導的，而且)(1)(yxf????(1)證明：證明：??xIx，給x以增量x?)0(xIxxx?????由)(xfy?在Ix上的單調性可知0)()(????

2、??xfxxfy于是yxxy?????1因直接函數(shù))(yx??在Iy上單調、可導，故它是連續(xù)的，且反函數(shù))(xfy?在Ix上也是連續(xù)的，當0??x時，必有0??y)(11limlim00yyxxyyx????????????即：)(1)(yxf????【例1】試證明下列基本導數(shù)公式().(arcsin)().()().(log)ln1112113122xxarctgxxaxax????????如果)(xu??在點x0可導，而)(ufy?

3、在點)(00xu??可導，則復合函數(shù)])([xfy??在點x0可導，且導數(shù)為)()(000xufdxdyxx??????證明：證明：因)(lim00ufxyu??????，由極限與無窮小的關系，有)00()(0????????????時當uuuufy用0??x去除上式兩邊得：xuxuufxy????????????)(0由)(xu??在x0的可導性有：00?????ux，0limlim00????????ux])([limlim000x

4、uxuufxyxx????????????????xuxuufxxx???????????????0000limlimlim)(?)()(00xuf?????即)()(000xufdxdyxx??????上述復合函數(shù)的求導法則可作更一般的敘述：若ux??()在開區(qū)間Ix可導，yfu?()在開區(qū)間Iu可導，且??xIx時，對應的uIu?，則復合函數(shù)])([xfy??在Ix內可導，且dxdududydxdy??(2)復合函數(shù)求導法則是一個非