作业帮ln[根号（x^2+y^2）] =arctany/x 求dy1/2*ln(x^2+y^2)=arctany/x两边对x求导,得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x-y)则dy=(x+y)/(x-y)*dx请问两边对求导是怎么得出1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/17 08:57:16
ln[根号（x^2+y^2）] =arctany/x 求dy1/2*ln(x^2+y^2)=arctany/x两边对x求导,得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x-y)则dy=(x+y)/(x-y)*dx请问两边对求导是怎么得出1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(
ln[根号（x^2+y^2）] =arctany/x 求dy
1/2*ln(x^2+y^2)=arctany/x两边对x求导,得
1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2
化简得
y'=(x+y)/(x-y)
则dy=(x+y)/(x-y)*dx
请问两边对求导是怎么得出1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2这个的?
还有这个我怎么看不太懂啊?
谁用数学软件如实的写出来就大谢了.
还有化解又是怎么得来的,本人小白.
请细致一些,不然看的我有压力啊
ln[根号（x^2+y^2）] =arctany/x 求dy1/2*ln(x^2+y^2)=arctany/x两边对x求导,得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x-y)则dy=(x+y)/(x-y)*dx请问两边对求导是怎么得出1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(
搞清复合函数依次求导.
z = ln√(x^2+y^2)
dz = [1/√(x^2+y^2)] d(√(x^2+y^2) )
=[1/√(x^2+y^2)] 1/[2√(x^2+y^2) ] d(x^2+y^2)
=[1/√(x^2+y^2)] 1/[2√(x^2+y^2) ] (2xdx+ 2ydy )
q= arctan(y/x)
dq = 1/( 1+ (y/x)^2 ) d (y/x)
=1/( 1+ (y/x)^2 ) (xdy -ydx) /x^2
去看看复合函数求导
ln(x^2+y^2)看做lnt，t=x^2+y^2，y=f（x）的三重复合函数
arctany/x看做arctant，t=y/x，y=f（x）