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已知z=arctany/x求az/ax,az/ay
人气:222 ℃ 时间:2020-01-27 12:28:50
解答
∂z/∂x
=1/(1+y^2/x^2)*(y/x)'
=x^2/(x^2+y^2)*y*(-1/x^2)
=-y/(x^2+y^2)
∂z/∂y
=1/(1+y^2/x^2)*(y/x)'
=x^2/(x^2+y^2)*(1/x)
=x/(x^2+y^2)求函数f(x)=2x^3-6x^2-18x-7的单调区间和极值
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