x^2+y^2=e^(arctan（y/x）),求dy/dx

x^2+y^2=e^(arctan（y/x）),求dy/dx
令F(x,y)=x^2+y^2-e^arctan(y/x)=0
對x求偏導(dǎo) 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)用題目中的x^2+y^2代替就可以把
即Fx=2x+y
對y求偏導(dǎo) 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)用題目中的x^2+y^2代替得 Fy=2y-x
∴ dy/dx= - Fx/Fy= - (2x+y) / (2y-x) = (2x+y)/(x-2y)

數(shù)學(xué)人氣：534 ℃時間：2019-10-11 04:01:40

優(yōu)質(zhì)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)這好像可以化為最簡吧(2x+y)/(x-2y)

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x^2+y^2=e^(arctan（y/x）),求dy/dx

精品偷拍一区二区三区,亚洲精品永久码,亚洲综合日韩精品欧美国产,亚洲国产日韩a在线亚洲