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u=ln(1/(x+√(y^2+z^2))),求u对x、u对y、u对z的偏导数,
是u=ln(x+√(y^2+z^2))。没有分之1。
人气:487 ℃ 时间:2020-06-15 09:49:58
解答
u=ln(1/(x+√(y^2+z^2)))du/dx=1/ 1/(x+√(y^2+z^2)) * -1/(x+√(y^2+z^2))^2 =-(x+√(y^2+z^2))dx/(x+√(y^2+z^2))^2=-1/(x+√(y^2+z^2))du/dy=1/ 1/(x+√(y^2+z^2)) * -1/(x+√(y^2+z^2))^2 *1/2√(y^2+z^...老大。。我问题写错了。。是u=ln(x+√(y^2+z^2))。。而且是求偏导数、、不是导数u=ln(x+√(y^2+z^2))du/dx=1/(x+√(y^2+z^2))du/dy=1/(x+√(y^2+z^2)) * 1/[2√(y^2+z^2) *2y =y/[(x+√(y^2+z^2))√(y^2+z^2)] du/dz=z/[(x+√(y^2+z^2))√(y^2+z^2)]
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