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Math Help - chain rule - second order - quick review

  1. #1
    Member billym's Avatar
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    chain rule - second order - quick review

    Let: f(u,v)=usin(v)

    where: u(x,y)=x-y and v(x,y)=e^{y-x}

    Use only the chain rule to find f_x, f_y, f_{xx}, f_{yy} in terms of v and u.

    So,

    f_u*u_x+f_v*v_x

    =sinv-e^{y-x}ucosv=sinv-vucosv

    f_y=f_u*u_y+f_v*v_y

    =-sinv+e^{y-x}ucosv=sinv+vucosv

    Correct?

    Then:

    Mostly wondering if this is the correct formula:

    f_{xx}=f_{uu}*u_{xx}+f_{vv}*v_{xx}

    In which case:

    f_{xx}=f_{yy}=(-usinv)e^{y-x}=-vusinv

    f_{xy}=(-usinv)(-e^{y-x})=vusinv
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  2. #2
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    Quote Originally Posted by billym View Post
    Let: f(u,v)=usin(v)

    where: u(x,y)=x-y and v(x,y)=e^{y-x}

    Use only the chain rule to find f_x, f_y, f_{xx}, f_{yy} in terms of v and u.

    So,

    f_u*u_x+f_v*v_x

    =sinv-e^{y-x}ucosv=sinv-vucosv

    f_y=f_u*u_y+f_v*v_y

    =-sinv+e^{y-x}ucosv=sinv+vucosv

    Correct?
    So far, so good!

    Then:

    Mostly wondering if this is the correct formula:

    f_{xx}=f_{uu}*u_{xx}+f_{vv}*v_{xx}
    No. f_{xx}= (f_x)_x= (f_u u_x+ f_v v_x)_x= (f_u)_x u_x+ f_u u_xx+ (f_v)_x v_x+ f_v v_{xx} = (f_{uu} u_x+ f_{uv} v_x)u_x+ f_u u_{xx}+ (f_{vu} u_x+ f_{vv} v_x)v_x+ f_v v_{xx}.

    In which case:

    f_{xx}=f_{yy}=(-usinv)e^{y-x}=-vusinv

    f_{xy}=(-usinv)(-e^{y-x})=vusinv
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  3. #3
    Member billym's Avatar
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    Red face

    f_u=sin(v), f_{uu}=0, f_v=ucos(v), f_{vv}=-usin(v), f_{uv}=f_{vu}=cos(v)

    u_x=1, u_{xx}=0, v_x=-v, v_{xx}=v

    u_y=-1, u_{yy}=0, v_y=v_{yy}=v, u_{xy}=0, v_{xy}=-v

    Have I made some calculation errors or are these totally wrong?

    f_{xx}=v[u(vsinv+cosv)-2cosv]

    f_{yy}=v[u(cosv-sinv)-2cosv]

    f_{xy}=-v[u(cosv-vsinv)-2cosv]

    (just a little tedious)
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