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0

I am showing the following question about the use of differential dl in section six of the user guide diffcoeff.

• Why does not the commanddl vec{x} work?


• When dealing with the unknown x, the numerical value between dl and x indicates the horizontal space of that number in mu before the x. But I do not think that happens in (4) and (5). Is it possible via a command that allows me to use dl vect{x} in the "natural way"? That is, it works (6).

Minimal working example

documentclass{article}

usepackage[ISO]{diffcoeff}



begin{document}



[

    diff{vec{x}}{vec{y}} % (1) Works.

    qquad

    vec{dl x} % (2) Works.

    quad

    {vec{}dl x} % (3) Works.

    quad

    dl5vec{x} % (4) Works.

    quad

    dl0vec{x} % (5) Works.

    quad

    %dl vec{x} % (6) Do not work.

]

end{document}

Thanks in advance.

Please add the tag [diffcoeff] in the question.

math-mode macros

share

asked 7 mins ago

OromionOromion

10218

add a comment | 

0

I am showing the following question about the use of differential dl in section six of the user guide diffcoeff.

• Why does not the commanddl vec{x} work?


• When dealing with the unknown x, the numerical value between dl and x indicates the horizontal space of that number in mu before the x. But I do not think that happens in (4) and (5). Is it possible via a command that allows me to use dl vect{x} in the "natural way"? That is, it works (6).

Minimal working example

documentclass{article}

usepackage[ISO]{diffcoeff}



begin{document}



[

    diff{vec{x}}{vec{y}} % (1) Works.

    qquad

    vec{dl x} % (2) Works.

    quad

    {vec{}dl x} % (3) Works.

    quad

    dl5vec{x} % (4) Works.

    quad

    dl0vec{x} % (5) Works.

    quad

    %dl vec{x} % (6) Do not work.

]

end{document}

Thanks in advance.

Please add the tag [diffcoeff] in the question.

math-mode macros

share

asked 7 mins ago

OromionOromion

10218

add a comment | 

I am showing the following question about the use of differential dl in section six of the user guide diffcoeff.

• Why does not the commanddl vec{x} work?


• When dealing with the unknown x, the numerical value between dl and x indicates the horizontal space of that number in mu before the x. But I do not think that happens in (4) and (5). Is it possible via a command that allows me to use dl vect{x} in the "natural way"? That is, it works (6).

Minimal working example

documentclass{article}

usepackage[ISO]{diffcoeff}



begin{document}



[

    diff{vec{x}}{vec{y}} % (1) Works.

    qquad

    vec{dl x} % (2) Works.

    quad

    {vec{}dl x} % (3) Works.

    quad

    dl5vec{x} % (4) Works.

    quad

    dl0vec{x} % (5) Works.

    quad

    %dl vec{x} % (6) Do not work.

]

end{document}

Thanks in advance.

Please add the tag [diffcoeff] in the question.

math-mode macros

share

asked 7 mins ago

OromionOromion

10218

documentclass{article}

usepackage[ISO]{diffcoeff}



begin{document}



[

    diff{vec{x}}{vec{y}} % (1) Works.

    qquad

    vec{dl x} % (2) Works.

    quad

    {vec{}dl x} % (3) Works.

    quad

    dl5vec{x} % (4) Works.

    quad

    dl0vec{x} % (5) Works.

    quad

    %dl vec{x} % (6) Do not work.

]

end{document}

share

asked 7 mins ago

OromionOromion

10218

share

asked 7 mins ago

OromionOromion

10218

asked 7 mins ago

OromionOromion

10218

asked 7 mins ago

OromionOromion

10218