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(cosy)^x=(sinx)^y求dy/dx
人气:393 ℃ 时间:2020-03-13 12:07:43
解答
(sinx)^y=(cosy)^x 两边取对数ln(sinx)^y=ln(cosy)^x yln(sinx)=xln(cosy)两边求导:y'ln(sinx) y/sinx*cosx=ln(cosy) x/cosy*(-siny)*y'y'ln(sinx) xy'/(sinycosy)=ln(cosy)-y/(sinxcosx)y'[ln(sinx) x/(sinycosy)]...
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