设z=(x2+y2)sec2(x+y),求

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问题 设z=(x2+y2)sec2(x+y),求

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答案由z=(x2+y2)sec2(x+y),得z=esec2(x+y)ln(x2+y2). 则[*]=esec2(x+y)ln(x2+y2)[2sec2(x+y)tan(x+y)ln(x2+y2)+[*]sec2(x+y)], [*]=esec2(x+y)ln(x2+y2)[2sec2(x+y)tan(x+y)ln(x2+y2)+[*]sec2(x+y)].

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