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求y=arctan根号[(1-x^2)/(1+x^2)]的微分
人气:249 ℃ 时间:2020-04-10 11:09:45
解答
y=arctan√[(1-x²)/(1+x²)]y'=1/[1+(1-x²)/(1+x²)]* 1/{2√[(1-x²)/(1+x²)]}* [-2x(1+x²)-2x(1-x²)]/(1+x²)²=-x/√[(1+x²)(1-x²)]
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