求函数的微分 第一题:y=(x^2-1)/(x^2+1) 求dy 第二题y=x乘以根号下(1-x^2)再乘以arcsinx,求dy
(sinx)的cosy次方=(cosx)的siny次方 求dy
人气:426 ℃ 时间:2019-08-18 07:20:37
解答
dy=(2x(x^2+1)-(x^2-1)2x)/(x^2+1)^2*dxe^(cosylnsinx)=e^(sinylncosx)cosylnsinx=sinylncosx-sinydylnsinx=cosycosxdx/sinx=cosydylncosx+siny(-sinx/cosx)dxdy=((1/tanx)+tanytanx)/(tanylnsinx+lncosx)*dx
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