How to use correctly dl with vectors using the diffcoeff package?
I am showing the following question about the use of differential dl in section six of the user guide diffcoeff.
- Why does not the command
dl vec{x}work? - When dealing with the unknown
x, the numerical value betweendlandxindicates 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 usedl 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
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 command
dl vec{x}work? - When dealing with the unknown
x, the numerical value betweendlandxindicates 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 usedl 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
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 command
dl vec{x}work? - When dealing with the unknown
x, the numerical value betweendlandxindicates 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 usedl 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
I am showing the following question about the use of differential dl in section six of the user guide diffcoeff.
- Why does not the command
dl vec{x}work? - When dealing with the unknown
x, the numerical value betweendlandxindicates 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 usedl 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
math-mode macros
asked 39 secs ago
OromionOromion
10218
10218
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