x^2+y^2=e^(arctan(y/x)),求dy/dx
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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)
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)
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)
再问: 这好像可以化为最简吧 (2x+y)/(x-2y)
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)
再问: 这好像可以化为最简吧 (2x+y)/(x-2y)
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