一个简单而又迷惑的求导问题

题目

一个简单而又迷惑的求导问题
已知：F = x + x*sin(3y) ,a = x*cos(y) ,b = x*sin(y)
求导：dF/da ,dF/db

答案

由链式法则,
∂F/∂a=(∂F/∂x)(∂x/∂a)+(∂F/∂y)(∂y/∂a)；
∂F/∂b=(∂F/∂x)(∂x/∂b)+(∂F/∂y)(∂y/∂b).
由F的表达式易知∂F/∂x=1+sin(3y),∂F/∂y=3x*cos(3y).
对a=x*cos(y)两边取全微分,得
(1) da=cos(y)dx-x*sin(y)dy；
对b=x*sin(y)两边取全微分,得
(2) db=sin(y)dx+x*cos(y)dy.
(1)*cos(y)+(2)*sin(y),得
cos(y)*da+sin(y)*db=[cos^2(y)+sin^2(y)]dx=dx,
所以∂x/∂a=cos(y),∂x/∂b=sin(y)；
(1)*sin(y)-(2)*cos(y),得
sin(y)*da-cos(y)*db=[-x*sin^2(y)-x*cos^2(y)]dy=-x*dy,
所以∂y/∂a=-sin(y)/x,∂y/∂b=cos(y)/x.
最后代入链式法则得
∂F/∂a=(1+sin(3y))(cos(y))+(3x*cos(3y))(-sin(y)/x)=cos(y)+cos(y)sin(3y)-3sin(y)cos(3y)；
∂F/∂b=(1+sin(3y))(sin(y))+(3x*cos(3y))(cos(y)/x)=sin(y)+sin(y)sin(3y)+3cos(y)cos(3y)