Bdtyktl

I am wondering if it is possible to add the extra symbols in this example given the way I have written the code. I have provided a minimal example and another image with the additional symbols of + and times.

documentclass{article}

usepackage{tikz,mathtools}

usetikzlibrary{tikzmark}

usetikzlibrary{decorations.pathreplacing}



%%% Derivative Macro



newcommand{der}[2]{dfrac{d#1}{d#2}}



%%% Derivative Prime Notation



newcommand{pder}[2]{#1^{prime}(#2)}





begin{document}



begin{align*}

 der{}{x} [tikzmarknode{A}{(}2x^{5}+x-1tikzmarknode{B}{)}tikzmarknode{C} 

 {(}3x-2tikzmarknode{D}{)}]&=tikzmarknode{E}{(}2x^{5}+x-1tikzmarknode{F} 

 {)}tikzmarknode[red]{G}{(}textcolor{red}{3}tikzmarknode[red]{H} 

 {)}+tikzmarknode{I}{(}3x-2tikzmarknode{J}{)}tikzmarknode[red]{K} 

 {(}textcolor{red}{10x^{4}+1}tikzmarknode[red]{L}{)}

 end{align*}



begin{tikzpicture}[overlay, remember picture]

draw [decoration={brace},decorate,thick,blue]

 ([yshift=5pt]A.north) -- node[midway, above=3pt] {$f(x)$} 

 ([yshift=5pt]B.north);

draw [decoration={brace},decorate,thick,blue]

 ([yshift=5pt]C.north) -- node[midway, above=3pt] {$g(x)$} 

  ([yshift=5pt]D.north);

 draw [decoration={brace},decorate,thick,blue]

  ([yshift=5pt]E.north) -- node[midway, above=3pt] {$f(x)$} 

  ([yshift=5pt]F.north);

 draw [decoration={brace},decorate,thick,red]

 ([yshift=5pt]G.north) -- node[midway, above=3pt] {$pder{g}{x}$} 

 ([yshift=5pt]H.north);

 draw [decoration={brace},decorate,thick,blue]

  ([yshift=5pt]I.north) -- node[midway, above=3pt] {$g(x)$} 

  ([yshift=5pt]J.north);

 draw [decoration={brace},decorate,thick,red]

 ([yshift=5pt]K.north) -- node[midway, above=3pt] {$pder{f}{x}$} 

 ([yshift=5pt]L.north);

 end{tikzpicture}



 end{document}

This outputs:

With the above braces I am trying add some additional operations: + and cdot

g'(x) is a little smaller in the code example but if I can add the symbols and I can also add space to compensate later. Is it possible???

I would actually like to argue that it is more ergonomic to define a style that just inserts the brace path with the node on top, and not to have so many scattered tikzmarknodes. This answer comes with an overbrace style, that can be used as

draw[blue,thick,overbrace={A with {$f(x)$} called f1}];

where A is the node you want to overbrace, {$f(x)$} is the node on top and f1 its name. Then it is also much easier to insert the additional signs.

documentclass{article}

usepackage{tikz,mathtools}

usetikzlibrary{tikzmark}

usetikzlibrary{decorations.pathreplacing,calc}



%%% Derivative Macro



newcommand{der}[2]{dfrac{mathrm{d}#1}{mathrm{d}#2}}



%%% Derivative Prime Notation



newcommand{pder}[2]{#1^{prime}(#2)}



tikzset{overbrace/.style args={#1 with #2 called #3}{

insert path={[decoration={brace},decorate] ([yshift=5pt]#1.north west)

--  node[midway, above=3pt] (#3) {#2}  

([yshift=5pt]#1.north east) }

}}



begin{document}



begin{align*}

 der{}{x} [tikzmarknode{A}{(2x^{5}+x-1)}

 tikzmarknode{B}{vphantom{x^{5}}(3x-2)}]

 &=tikzmarknode{C}{(2x^{5}+x-1)}

 tikzmarknode[red]{D}{vphantom{x^{5}}(3H)}

 +tikzmarknode{E}{vphantom{x^{5}}(3x-2)}

 tikzmarknode[red]{F}{(10x^{4}+1)}

 end{align*}



begin{tikzpicture}[overlay, remember picture]

draw[blue,thick,overbrace={A with {$f(x)$} called f1}];

draw[blue,thick,overbrace={B with {$g(x)$} called g1}];

draw[blue,thick,overbrace={C with {$f(x)$} called f2}];

draw[red,thick,overbrace={D with {$pder{g}{x}$} called dg1}];

draw[blue,thick,overbrace={E with {$g(x)$} called g2}];

draw[red,thick,overbrace={F with {$pder{f}{x}$} called df1}];

path ($(A.east)!0.5!(B.west)$) coordinate (auxAB)

($(C.east)!0.5!(D.west)$) coordinate (auxCD)

($(D.east)!0.5!(E.west)$) coordinate (auxDE)

($(E.east)!0.5!(F.west)$) coordinate (auxEF);

path (auxAB|-f1) node[blue]{$cdot$}

([xshift=-3pt]auxCD|-f1) node[blue]{$cdot$} %manual correction

(auxDE|-f1) node[blue]{$+$}

(auxEF|-f1) node[blue]{$cdot$};

end{tikzpicture}



end{document}

If you want to layout of the braces, you only need to adjust the style. For instance, if you want gaps between them, you could do

tikzset{overbrace/.style args={#1 with #2 called #3}{

insert path={[decoration={brace},decorate] ([yshift=5pt,xshift=1pt]#1.north west)

--  node[midway, above=3pt] (#3) {#2}  

([yshift=5pt,xshift=-1pt]#1.north east) }

}}

which yields

Note also that I put the signs on top of the signs of the underlying equation. If you want them in the middle, this will be even easier.

documentclass{article}

usepackage{tikz,mathtools}

usetikzlibrary{tikzmark}

usetikzlibrary{decorations.pathreplacing}



%%% Derivative Macro



newcommand{der}[2]{dfrac{mathrm{d}#1}{mathrm{d}#2}}



%%% Derivative Prime Notation



newcommand{pder}[2]{#1^{prime}(#2)}



tikzset{overbrace/.style args={#1 with #2 called #3}{

insert path={[decoration={brace},decorate] ([yshift=5pt,xshift=1pt]#1.north west)

--  node[midway, above=3pt] (#3) {#2}  

([yshift=5pt,xshift=-1pt]#1.north east) }

}}



begin{document}



begin{align*}

 der{}{x} [tikzmarknode{A}{(2x^{5}+x-1)}

 tikzmarknode{B}{vphantom{x^{5}}(3x-2)}]

 &=tikzmarknode{C}{(2x^{5}+x-1)}

 tikzmarknode[red]{D}{vphantom{x^{5}}(3)}

 +tikzmarknode{E}{vphantom{x^{5}}(3x-2)}

 tikzmarknode[red]{F}{(10x^{4}+1)}

 end{align*}



begin{tikzpicture}[overlay, remember picture]

draw[blue,thick,overbrace={A with {$f(x)$} called f1}];

draw[blue,thick,overbrace={B with {$g(x)$} called g1}];

draw[blue,thick,overbrace={C with {$f(x)$} called f2}];

draw[red,thick,overbrace={D with {$pder{g}{x}$} called dg1}];

draw[blue,thick,overbrace={E with {$g(x)$} called g2}];

draw[red,thick,overbrace={F with {$pder{f}{x}$} called df1}];

path[blue] (f1) -- (g1) node[midway] {$cdot$}

(f2) -- (dg1)  node[midway] {$cdot$}

(dg1) -- (g2)  node[midway] {$+$}

(g2) -- (df1)  node[midway] {$cdot$};

end{tikzpicture}



end{document}

Despite being a very powerful tool, tikz can be too costly for some tasks. Here I use overbrace and a small hack to add the operators while occupying 0pt width using a makebox[0pt]{..} macro. Also, as @LoopSpace suggested, we can make the braces at the same height by adding vphantom{x^5} to parentheses without that term.

documentclass{article}

usepackage{amsmath}

begin{document}



newcommand{hide}[2]{makebox[0pt][l]{$hspace{#1pt}#2$}}



[

frac{d}{dx}[overbrace{(2x^5+x-1)}^{f(x)hide{20}{cdot}} overbrace{(vphantom{x^5}3x-2)}^{g(x)}]

= overbrace{(2x^5+x-1)}^{f(x)hide{15}{cdot}} overbrace{(vphantom{x^5}3)}^{g'(x)hide{2}{+}} 

+ overbrace{(vphantom{x^5}3x-2)}^{g(x)hide{10}{cdot}} overbrace{(10x^4+1)}^{f'(x)}

]



end{document}

For fun, another simple solution with the overbracecommand and pstricks:

documentclass{article}

usepackage{mathtools, esdiff}

usepackage{pst-node}



begin{document}



begin{align*}

 diff{}{x}bigl[(textcolor{red}{overbrace{strutcolor{black}2x^{5}+x-1}^{Rnode{f}{textstyle f(x)}} textcolor{black}{ )(}overbrace{strutcolor{black}3x-2}^{Rnode{g}{textstyle g(x)}}})bigr] &= (textcolor{blue}{overbrace{color{black}2x^{5}+x-1}^{Rnode{bluef}{textstyle f(x})}})textcolor{red}{overbrace{strut(3)}^{Rnode{dg}{textstyle g'(x)}} }+ (textcolor{blue}{overbrace{strutcolor{black}3x-2}^{Rnode{blueg}{textstyle g(x)}}} )textcolor{red}{(overbrace{10x^{4}+1}^{Rnode{df}{textstyle f'(x )}})}%

 %% Additional symbols

 psset{linestyle=none}

 ncline{f}{g}ncput{boldsymbolcdot}

 ncline{bluef}{dg}ncput{boldsymbolcdot}

 ncline{blueg}{df}ncput{boldsymbolcdot}

 ncline{blueg}{dg}ncput{ + }

 end{align*}



 end{document}