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tan x y 的二阶导

y'= dy/dx =sec^2(x+y)(1+y'); →[sec^2(x+y) -1]y'=sec^2(x+y); →[tan^2(x+y) ]y'=sec^2(x+y); →y'=1/sin^2(x+y); 则:y'' =dy' /dx=d[sin^(-2)(x+y)] /dx=(-2)sin^(-3)(x+y) cos(x+y)(1+y') =-2sin^(-3)(x+y) cos(x+y)[1+sin^(-2)(x+y)]=-2cos(x+y)[sin^(-3)(x+y) +sin^(-5)(x+y)]

y=tan(x+y)y'=sec(x+y)*(x+y)'=sec(x+y)*(1+y')=sec(x+y)+y'sec(x+y)y'-y'sec(x+y)=sec(x+y)y'=sec(x+y)/[1-sec(x+y)]=sec(x+y)/{-[sec(x+y)-1]}=sec(x+y)/[-tan

两边对x求导得:[sec(x+y)](1+y')=y'解得:y'=sec(x+y)/[1-sec(x+y)]=-sec(x+y)/tan(x+y)=-csc(x+y)再求导得:y''=-2csc(x+y)[-csc(x+y)cot(x+y)](1+y')=2(1+y')csc(x+y)cot(x+y)将y'=-csc(x+y)代入得:y''=2[1-csc(x+y)]csc(x+y)cot(x+y)=-2csc(x+y)cot(x+y)【数学之美】团队为您解答,若有不懂请追问,如果满意请点下面的“选为满意答案”.

解:两边对x求导,得 y'=sec^2 (x+y)*(1+y')=(1+y^2)(1+y') 解得y'=-(1+y^2)/y^2=-1-y^(-2) 两边再对x求导,得 y"=2y^(-3)*y'=-2(1+y^2)/y^5 不明白请追问.

解答:百y=tan(x-y)两边求导y&度#39; = (1-y')/cos(x-y)y' = 1/[1+cos(x-y)]再次求导y" = -2cos(x-y)[-sin(x-y)](1-y')/[1+cos(x-y)]= sin(2x-2y)(1-y')/[1+cos(x-y)]= sin(2x-2y){1 - 1/[1+cos(x-y)]}/[1+cos(x-y)]= sin(2x-2y)cos(x-y)/[1+cos(x-y)]

y=tan(x-y)两边求导y' = (1-y')/cos(x-y)y' = 1/[1+cos(x-y)]再次求导y" = -2cos(x-y)[-sin(x-y)](1-y')/[1+cos(x-y)]= sin(2x-2y)(1-y')/[1+cos(x-y)]= sin(2x-2y){1 - 1/[1+cos(x-y)]}/[1+cos(x-y)]= sin(2x-2y)cos(x-y)/[1+cos(x-y)]

y'=(1+y')sec^2(x+y)y'=sec^2(x+y)/(1-sec^2(x+y))=1+cot^2(x+y))y''=-(1+y')cot(x+y)csc^(x+y)=-(2+cot^2(x+y))cot(x+y)csc^(x+y)

y=tan(x+y)y'=tan'(x+y)=sec^2(x+y)(x+y)'=sec^2(x+y)*(1+y')y'=sec^2(x+y)/[1-sec^2(x+y))=-sec^2(x+y)/tan^2(x+y)=-1/sin^2(x+y)=-csc^2(x+y)y''=-2csc(x+y)*[csc(x+y)]'=-2csc(x+y)*[-csc(x+y)cot(x+y)](x+y)'=2csc^2(x+y)cot(x+y)(1+y')=2csc^2(x+y)cot(x+y)[1-csc^2(x+y)].1-csc^2(x+y)=cot^2(x+y)

y=tan(x+y)两边求导:y'=(sec(x+y))^2*(1+y'),所以y'=-(csc(x+y))^2=-1-1/y^2所以,y''=2y'/y^3=-2(1+y^2)/y^5

y'=[sec(x+y)]^2 * y' y''=2sec(x+y)sec(x+y)tan(x+y)(y')^2+y''[sec(x+y)]^2 y''={2sec(x+y)sec(x+y)tan(x+y)(y')^2}/{1-[sec(x+y)]^2}

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