x^2+y^2=e^(arctan（y/x）),求dy/dx令F(x,y)=x^2+y^2-e^arctan(y/x)=0对x求偏_数学解答

x^2+y^2=e^(arctan（y/x）),求dy/dx
令F(x,y)=x^2+y^2-e^arctan(y/x)=0
对x求偏导 Fx = 2x-e^arctan(y/x) * 1/[1+(y/x)^2] * [-y/(x^2)]=2x + y/(x^2+y^2) * e^arctan(y/x)
然后e^arctan(y/x)用题目中的x^2+y^2代替就可以把
即Fx=2x+y
对y求偏导 Fy=2y-e^arctan(y/x)*1/[1+(y/x)^2]*(1/x)=2y-x/(x^2+y^2) * e^arctan(y/x)
同上把e^arctan(y/x)用题目中的x^2+y^2代替得 Fy=2y-x
∴ dy/dx= - Fx/Fy= - (2x+y) / (2y-x) = (2x+y)/(x-2y)

人气：135 ℃　时间：2019-10-11 04:01:40

解答

x^2+y^2-e^(arctan（y/x）)=0
2x+2y*y'-(arctan（y/x）)'e^(arctan（y/x）)=0
2x+2y*y'-1/(1+y^2/x^2)*(y'x-y)/x^2 *e^(arctan（y/x）)=0
2x+2y*y'-(y'x-y)/(x^2+y^2)*e^(arctan（y/x）)=0
2x+2y*y'-y'x/(x^2+y^2)*e^(arctan（y/x）)+y/(x^2+y^2)*e^(arctan（y/x）)=0
y'=(2x++y/(x^2+y^2)*e^(arctan（y/x）))/(x/(x^2+y^2)*e^(arctan（y/x）)-2y)这好像可以化为最简吧(2x+y)/(x-2y)