设y=(tan2x)^cot(x/2) ,求dy/dx

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设y=(tan2x)^cot(x/2) ,求dy/dx
设y=(tan2x)^cot(x/2) ,求dy/dx

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

y'=y{-[csc(x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x

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
lny = cot(x/2)ln(tan2x)
两边关于x求导,y'/y=[cot(x/2)]'ln(tan2x) + cot(x/2)[ln(tan2x)]'
y'/y=[-(cscx)^2/2]ln(tan2x)+2cot(x/2)(secx)^2/tan(2x)
所以y'=y{[-(cscx)^2/2]ln(tan2x)+2cot(x/2)(secx)^2/tan(2x)}
y'= dy/dx =(tan2x)^cot(x/2)[-(cscx)^(2)ln(tan2x)/2+2cot(x/2)(secx)^2/tan(2x)]