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tAn x y 二次求导

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)]

[tan(x/2)]'=sec(x/2)*(x/2)'=sec(x/2)*(1/2)=(1/2)sec(x/2)

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(x+y)]=-1/cos(x+y)*cos(x+y)/sin(x+y)=-csc(x+y) y''=-2csc(x+y)*[

tan(x^2)=sin(x^2)/cos(x^2)利用分数的求导公式(u/v)'=(u'v-uv')/v^2可知原式的导等于[2x(sin(x^2))^2+2x(cos(x^2))^2]/(cos(x^2))^2因为sin^2+cos^2=1 所以=2x/(cos(x^2))^2 或者写成正割的形式 就是2x(sec(x^2))^2

二阶导数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(x+y)]=-1/cos(x+y)*cos(x+y)/sin(x+y)=-csc(x+y)y''=-2

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)两边同时对x进行求导:y'=sec(x+y)*(1+y') [1-sec(x+y)]y'=sec(x+y) -tan(x+y)y'=sec(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=tan(x/y) y'=sec(x/y)(y-xy')/y ycos(x/y)y'+xy'=y y'=y/[ycos(x/y)+x]

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