求复合函数偏导数链式法则解.ppt

§2.求复合函数偏导数的链式法则 链式法则 定理1设n=f(x,y),x=φ(x,),y=v(x,),此时, 若 ax ay ax ay 及 atat as a 在某点(s)都存在,而f(x,y) 在相应于(,)的点(x,y)可微,则成立公式: ou at 及 OuOu ax au ay as ax as y University 函v §2.求复合函数偏导数的链式法则 该公式称为求复合函数偏导数的链式法则 链式法则如图示 ou au ax au oy at ax at ay a auau ax a as ar as ay a y University 函v §2.求复合函数偏导数的链式法则 特殊情形 (1)=f(x,y,x=q(t),y=v(t),则有 du au dx au dy 十 dt ar dt ay dt (2)=f(x,y,1,x=q(s,t),y=v(s,),则有 ou au ax ou ay ou at ax at ay at (at y University 函v §2.求复合函数偏导数的链式法则 auau ax au 十 as ax as ay a 例1设u=x2+y,x=2st,y=t+s2,求Oat as at 解 2x·2t+1.2s=4xt+2s Ou au ax au ay =2x·2s+1.s=4xs+1 at aa ay a y University 函v §2.求复合函数偏导数的链式法则 例2设z=f(x,y),y=(於女d 解:a=ax+ dx ar ar ay a fr+f,o(r) 再求导,得 d2z. Ofxx ar ar l ar+ ay(x)o(x)+f, o(r) afr Oy a, af dr ∫+2fnq(x)+∫[φ(x)]+∫,卯(x) y University 函v