y=loge(x^2+y^2),那么dy/dx在(1,0)的值是?

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y=loge(x^2+y^2),那么dy/dx在(1,0)的值是?
y=loge(x^2+y^2),那么dy/dx在(1,0)的值是?

y=loge(x^2+y^2),那么dy/dx在(1,0)的值是?
dy/dx=1/(x^2+y^2) *[2x+2ydy/dx]
代入(1,0)可知道
dy/dx=1/1*[2+0]=2

y' = 2x/(x^2+y^2) + 2y'y/(x^2 + y^2)
在(1,0)点
y'=2

根据题意有：
y=ln(x^2+y^2)
所以：
y'=(x^2+y^2)'/(x^2+y^2)
=(2x+2yy')/(x^2+y^2)
所以：
y'=2x/(x^2+y^2-2y)
y'(1,0)
=2/(1+0)
=2.

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