设y=(tan2x)^cot(x/2) ,求dy/dx
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设y=(tan2x)^cot(x/2) ,求dy/dx
楼上好像写错了,要细心啊
两边取对数,得
lny=ln【(tan2x)^cot(x/2) 】=cot(x/2)ln(tan2x)
两边再分别求导,得
y'/y={-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x
即
y'=y{-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x
有
y'=dy/dx =(tan2x)^cot(x/2)*{-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)*(sec2x)^2}/tan2x
两边取对数,得
lny=ln【(tan2x)^cot(x/2) 】=cot(x/2)ln(tan2x)
两边再分别求导,得
y'/y={-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x
即
y'=y{-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x
有
y'=dy/dx =(tan2x)^cot(x/2)*{-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)*(sec2x)^2}/tan2x
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