arctan x/1+y2,x=1,y=1全微分

来源：学生作业学帮网编辑：学帮网时间：2024/06/18 01:55:37

z=arctan[x/(1+y²)]
∂z/∂x={1/[1+x²/(1+y²)]}*[1/(1+y²)]
∂z/∂y={1/[1+x²/(1+y²)]}*[-x/(1+y²)²]*2y
所以dz={1/[1+x²/(1+y²)]}*[1/(1+y²)]dx+{1/[1+x²/(1+y²)]}*[-x/(1+y²)²]*2ydy
x=1,y=1
dz=(2/5)dx-(2/5)dy

arctan x/1+y2,x=1,y=1全微分 arctan(y)=x+1, y=? y=arctan(1/x)求导 y=arctan（x^2+1） y=arctan|x|y=In(1+x) y=arctan(1-x)/(1+x)的微分 y=arctan(x+1/x-1)导数求y=arctan(x+1／x-1)的导数函数求导.y=arctan(x+1)/(x-1) y=arctan(1-x/1+x),求dy 求导y=arctan[2x/(1-x^2)] 急等求导 Y=ARCTAN x/1+x^2 y=(x+1)arctan√x 求dy 设 y=4 arctan x ,则y'(1)=? 请问y=arctan(1-x),求y' y=ln[arctan(1-x)],这dy= z=arctan x/(1+y^2),则dz=? 设y=arctan 1/x,则dy=rt