(cosx-xsinx)/(sinx+xcosx)求导

1个回答

  • [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