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y=e^[-sin^2(1/x)]的微分
人气:235 ℃ 时间:2020-10-01 09:53:26
解答
dy=d{e^[-sin^2(1/x)]} =e^[-sin^2(1/x)]*[-2sin(1/x)]*cos(1/x)*(-1/x²)dx =e^[-sin^2(1/x)]2sin(1/x)cos(1/x)*(1/x²)dx=e^[-sin^2(1/x)] *sin(2/x) *(1/x²)dx
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