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求y=arctan[2x/(1-x^2)]的导数,

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人气:197 ℃ 时间:2020-05-02 10:01:48
解答
y=arctan[2x/(1-x^2)] y=arctanu u=2x/(1-x^2) u'=(2(1-x^2)-2x(-2x))/(1-x^2)^2=(2x^2+2)/(1-x^2)^2那么导数 y'=1/(1+u^2)*u'=1/(1+4x^2/(1-x^2)^2) * ( 2x^2+2)/(1-x^2)^2=( 2x^2+2)/((1-x^2)^2+4x^2) 底下是个完...
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