(cosx-xsinx)/(sinx+xcosx)求导这个是dy/dx 参数方程是 x=tsint+2 求 dy/dx,d^2y/dx^2y=2+tcost

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(cosx-xsinx)/(sinx+xcosx)求导这个是dy/dx参数方程是x=tsint+2求dy/dx,d^2y/dx^2y=2+tcost(cosx-xsinx)/(sinx+xcosx)求

(cosx-xsinx)/(sinx+xcosx)求导这个是dy/dx 参数方程是 x=tsint+2 求 dy/dx,d^2y/dx^2y=2+tcost
(cosx-xsinx)/(sinx+xcosx)求导
这个是dy/dx 参数方程是 x=tsint+2 求 dy/dx,d^2y/dx^2
y=2+tcost

(cosx-xsinx)/(sinx+xcosx)求导这个是dy/dx 参数方程是 x=tsint+2 求 dy/dx,d^2y/dx^2y=2+tcost
[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/g^2(x)
分子求导:
[(cosx-xsinx)]'=-sinx-[sinx+xcosx]=-2sinx-xcosx
分母求导:
[(sinx+xcosx)]'=cosx+[cosx-xsinx]=2 cosx-xsinx
[(cosx-xsinx)/(sinx+xcosx)]'
=[(-2sinx-xcosx)*(sinx+xcosx)-(cosx-xsinx)*(2cosx-xsinx)]/(sinx+xcosx)^2
=(-2-x^2)/(sinx+xcosx)^2

参数方程是
x=tsint+2 y=2+tcost
求 dy/dx, d^2y/dx^2
x'=sint+tcost
y'=cost-tsint
x''=cost+cost-tsint=2cost-tsint
y''=-sint-sint-tcost=-2sint-tcost
dy/dt=(2+tcost)'/(tsint+2)'
=(cost-tsint)/(sint+tcost)
d^2y/dx^2=d/dx*(dy/dx)
=[(-2sint-tcost)(sint+tcost)-(cost-tsint)(2cost-tsint)]/(sint+tcost)^3
=(-2-t^2)/(sint+tcost)^3

X->0 (sinx)^2/(1-cosx+xsinx) 的极限 已知f(x)=根号3sinx-cosx,xsinx=4/5求f(x) 求当X趋近于0 (sinx)平方 / (1-cosx+xsinx)的极限 (1-coax-xsinx)/(2-2cosx-sinx^2)当x趋于0时的极限 函数(xsinx+cosx)/x^2 - cosx/(sinx)^2 x趋向于0 的极限是不是1呀 xcosx-sinx怎么变成cosx-xsinx-cosx的? 请帮我写[(sinX+XcosX)(1+cosX)+(XsinX)^2]/(1+cosX)^2得出结果的详细过程谢谢!这个题传说中结果应该是:(sinX+X)/(1+cosX) 证明当x→0时,[√(1+xsinx)-√(cosx)]~(3/4)x^2.当x→0时,[√(1+xsinx)-√(cosx)]~(3/4)x^2 lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2] =lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]} =(2/3)lim(1+xsinx-cosx)/(x^2) =(2/3)lim(sinx+xcosx+si .证明当x→0时,[√(1+xsinx)-√(cosx)]~(3/4)x^2.当x→0时,[√(1+xsinx)-√(cosx)]~(3/4)x^2 lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2] =lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]} =(2/3)lim(1+xsinx-cosx)/(x^2) =(2/3)lim(sinx+xcosx+s 1/x/sinx/1-cosx=1-cosx/xsinx=1/2 这个是怎么变的啊原题是lnx/ln(1-cosx) lim趋于0+ y=(sinx-xcosx)/(cosx+xsinx)的导数 求y=sinx/cosx+xsinx 的导函数RT 求不定积分∫sinx-xcosx/cosx+xsinx dx limx趋于0 (sinx)^2/ (1+xsinx+√cosx) (-xsinx-cosx)/(x^2)=(-xcosx+cosx)/(x^2)? y=lim (x → 0) ( √1+xsinx - √cosx) / arcsin^2x=lim (x → 0) ( 1+xsinx - cosx) / x^2( √1+xsinx +√cosx) =1/2 lim (x → 0) 1-cosx /x^2+ sinx/x这最后那步是怎么得到的,麻烦说的详细点,描述公式, y=lim (x → 0) ( √1+xsinx - √cosx) / arcsin^2x=lim (x → 0) ( 1+xsinx - cosx) / x^2( √1+xsinx +√cosx) =1/2 lim (x → 0) 1-cosx /x^2+ sinx/x这两步是怎么得到的,麻烦说的详细点,描述公式, 设函数:y=xsinx^2,求dy/dx.我算的是=sinx^2+x2sinxcosx,但是答案是:sinx^2+2x^2cosx^2