【文章內(nèi)容簡介】
)1s i nln( c o s )ln( s i n)(lns i nlns i nlns i nxxxxexxeeyxxxxxx????????)s i nln( c o ss i n x xxxx x ???xxy s in?的導(dǎo)數(shù) 方法,求出然后利用復(fù)合函數(shù)求導(dǎo)yxxey lns in ??轉(zhuǎn)化為指數(shù)函數(shù)例 5 .,)4)(3()2)(1( yxxxxy ?????? 求設(shè)解 等式兩邊取對(duì)數(shù)得 )]4l n ()3l n ()2l n ()1[ l n (21ln ???????? xxxxy求導(dǎo)得上式兩邊對(duì) x]4131)2( 111[21 ????????? xxxxyy]4131)2( 111[)4)(3( )2)(1(21 ????????? ???? xxxxxx xxy)( tx ??設(shè)函數(shù) ,)()(中在參數(shù)方程?????tytx??),(1 xt ?? ?具有單調(diào)連續(xù)的反函數(shù))]([ 1 xy ??? ??,0)()(),( ??? ttytx ??? 且都可導(dǎo),再設(shè)函數(shù),也稱參數(shù)方程為參數(shù)式函數(shù) . 方法一 、分別對(duì) yx, 求微分 ,通過微商得到導(dǎo)數(shù) . .)()()( ,)(ttdtdydttdydttdx???????????三 參數(shù)方程確定函數(shù)的導(dǎo)數(shù) dxdtdtdydxdy ??dtdxdtdy 1??)()(tt?????由復(fù)合函數(shù)及反