Derivative of a product

Question

In calculus, the derivative of a mathematical function defines the rate at which it changes. The derivative of a function f(x) can be marked as f'(x), and these can sometimes be abbreviated to f and f'.

The derivative of the product of two functions fg is f'g + g'f. The derivative of three fgh is f'gh + fg'h + fgh'. In general, the derivative of a product of any number of functions is the sum of the product of all but one, multiplied by the derivative of the remaining one, for each individual function.

Your challenge is to take a string of distinct alphabetical characters and transform it into its derivative. No spacing or simplification is required, and terms separated by + may be in any order. The string will contain at least two characters.

This is code-golf, so the shortest code wins!

Testcases

abc -> a'bc+ab'c+abc'
de -> d'e+de'
longabcdef -> l'ongabcdef+lo'ngabcdef+lon'gabcdef+long'abcdef+longa'bcdef+longab'cdef+longabc'def+longabcd'ef+longabcde'f+longabcdef'
short -> s'hort+sh'ort+sho'rt+shor't+short'

Answer 1

Pip, 15 bytes

aRL#aJ'+<>#aJ''

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Explanation

aRL#aJ'+<>#aJ''
a                Command-line arg       "abc"
   #a            Length of arg          3
 RL              Repeat-list            ["abc"; "abc"; "abc"]
     J'+         Join on "+"            "abc+abc+abc"
        <>#a     Groups of size len(a)  ["abc"; "+ab"; "c+a"; "bc"]
            J''  Join on "'"            "abc'+ab'c+a'bc"

Answer 2

Python, 54 bytes

lambda s:((l:=len(s))*(l*"'"+-~l*(s+"+")))[-~l*~l::-l]

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Python, 56 bytes

lambda s:((l:=-~len(s))*(l*(s+"+")+-~l*"'"))[l:-3*l:-~l]

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Old Python, 58 bytes

lambda s:((l:=-~len(s))*(l*(s+"+")+l*"'"+"+"))[l:-3*l:-~l]

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How?

 a b c d +
(a)b c d +
 a(b)c d +
 a b(c)d +
 a b c(d)+
 ' ' ' '(')'
 a b c d(+)
 a b c d +
(a)b c d +
 a(b)c d +
 a b(c)d +
 ' ' '(')' '
 a b c(d)+
 a b c d(+)
 a b c d +
(a)b c d +
 a(b)c d +
 ' '(')' ' '
 a b(c)d +

   etc.

Answer 3

C (clang), 69 64 59 bytes

-10 thanks to @Noodle9 and @ceilingcat

l;f(char*a){for(l=0;a[l];)printf("+%.*s'%s"+!l,++l,a,a+l);}

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Answer 4

Brainfuck, 87 84 bytes

>,>---[>+<+++++++]>++[>,]<[<]<[<]>[[.>]>.>[.[-<<+>>]>[.>]<[<]<++++.----[->+<]<[<]]>]

Saved 3 bytes thanks to Nitrodon.

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Explanation:

memory progress for abc:
null a null 'bc
null ab null 'c
null abc' null

string is split to two parts by null character

>,>                load first character
---[>+<+++++++]>++ load ' to memory; taken form esolangs wiki
[>,]               load rest of input
<[<]<[<]>          go to beginning of memory
[ 
 [.>]              print first part
 >.>               print '
 [                 check if there is text after '
  .                print first char of second part 
  [-<<+>>]         move char to the first part
  >[.>]            print rest of second part
  <[<]<            go to position of '
  ++++.----        use ' to print plus
  [->+<]           move '
  <[<]             go before beginning
 ]
 >                 move back to beginning
                   if there wasn't text after ' move after end instead
]

Answer 5

Python, 47 bytes

lambda s:"+".join(s.replace(c,c+"'")for c in s)

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How it works:

a  b  c  d --> a' b  c  d 
a  b  c  d --> a  b' c  d 
a  b  c  d --> a  b  c' d 
a  b  c  d --> a  b  c  d'

then join with '+'

Answer 6

Factor, 53 bytes

[ dup length [1,b] [ cut "'"glue ] with map "+"join ]

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If the order of each product doesn't matter:

Factor, 43 bytes

[ all-rotations [ 39 suffix ] map "+"join ]

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Explanation

                  ! "short"
all-rotations     ! { "short" "horts" "ortsh" "rtsho" "tshor" }
[ 39 suffix ] map ! { "short'" "horts'" "ortsh'" "rtsho'" "tshor'" }
"+"join           ! "short'+horts'+ortsh'+rtsho'+tshor'"

Answer 7

JavaScript (Node.js), 46 44 bytes

l=>[...l].map(s=>l.replace(s,s+"'")).join`+`

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-2 bytes thanks to emanresu A

Answer 8

Jelly, 14 bytes

J‘œṖ€j€”'j”+”'

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J‘œṖ€j€”'j”+”'  Full Program
J               [1, 2, ..., len]
 ‘              [2, 3, ..., len + 1]
  œṖ€           For each of these, partition (*)
     j€”'       Join each partition on ' (**)
         j”+    Join on +
            ”'  An unparseable nilad forces the current string to be printed

(And then the ' is printed at the end)

(*)  partition takes a list on the left but since we give it a single number,
     it auto-wraps it so we just get all needed splits
(**) the last partition only has one sublist so join doesn't work properly here

Answer 9

Retina, 15 13 bytes

L$|+`.
$>`'$'

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-2 thanks to Neil.

Match every character. For each one, take the text before the separator after the matched string, followed by an apostrophe, followed by the text after the matched string. Print as a list with separator +.

Answer 10

K (ngn/k), 21 bytes

Most straightforward K solution.

{"+"/?[x;;"'"]'1+!#x}

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Explanation

{"+"/?[x;;"'"]'1+!#x}
     ?[x;;"'"]          / insert "'" in string at 
              '         / each position
               1+!#x    / 1..len of string
 "+"/                   / join the resulting strings with "+"

Answer 11

Wolfram Language (Mathematica), 45 36 bytes

aRiffle[a/.#->#<>"'"&/@a,"+"]<>""

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Input [{characters...}]. Assumes all characters in input are unique. Returns a string.

---

Wolfram Language (Mathematica), 35 bytes

ToString@Expand[Dt[1##]/._@a_:>a']&

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Input [symbols...]. Assumes symbols are unique and not defined. Returns a spaced string.

Dt treats strings as constants.

Answer 12

JavaScript (Node.js), ^{83 \$\cdots\$ 72} 69 bytes

s=>g=(r="",i=0)=>s[i]?g(r+("+"+s).slice(!i,++i+1)+"'"+s.slice(i),i):r

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Saved 7 8 bytes thanks to emanresu A!!!
Saved 3 bytes thanks to Samathingamajig!!!

Answer 13

Charcoal, 10 bytes

⪫⪪⪫Ｅθθ+Ｌθ'

Try it online! Link is to verbose version of code. Explanation: Port of @DLosc's Pip answer.

    θ       Input string
   Ｅ        For each character
     θ      Input string
  ⪫         Join on
      +     Literal string `+`
 ⪪          Split into substrings of length
        θ   Input string
       Ｌ    Length
⪫           Join on
         '  Literal string `'`
            Implicitly print

Answer 14

Husk, 14 12 bytes

J''CL¹J'+SRL

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Same approach as DLosc's Pip answer.

Answer 15

BQN, 21 bytes^SBCS

1↓⟜∾'+'∾¨1↓↑∾⟜'''⊸∾¨↓

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↑ gives prefixes, ↓ suffixes. ∾⟜'''⊸∾¨ joins matching prefixes and suffixes with a single quote.

1↓ removes the first value (empty prefix ∾ ' ∾ full suffix).

'+'∾¨ prepends a + to each string.

1↓⟜∾ joins the string and removes the leading +.

Answer 16

BQN, 28 bytes^SBCS

{1↓⥊'''∾˘⟨∘,≠𝕩⟩⥊«⥊'+'∾˘⊢⌜˜𝕩}

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returns the required string with a trailing space. removing that is +3 bytes

-3 from ovs, who also made this tacit.

Answer 17

Bash, 83 bytes

v=$1\';until [ $v = \'$1 ];do a+=+$v;v=$(sed "s/\(.\)'/'\1/"<<<$v);done;echo ${a#+}

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Answer 18

C (gcc), 151 bytes

char*f(char*c,int i){char*r=malloc(i*i+i);char*o=r,k=0;for(;k<i*i;){if(k++%(i+1)==1)*o++='\'';*o++=*c++;if(!*c)c-=i,*o++='+';}*o--=0;*o='\'';return r;}

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Answer 19

Julia, 40 bytes

a->join([replace(a,i=>"$i'") for i=a],+)

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only works with all letters distinct

Why...how...what...

• [... for i=a] iterates over the string a and store the results in a list. i will be a Char
• replace(a, i=>"$i'") replaces all characters equal to i in a with i followed by '
• join(..., +) joins the list with +. + is a function, so string(+) == "+" is used.

Answer 20

R, 81 75 bytes

Edit: -6 bytes thanks to Giuseppe

function(x,`[`=substring)paste0(x[1,s<-1:nchar(x)],"'",x[s+1],collapse="+")

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Takes advantage of the vectorization of substring over a vector of start (& optional end) positions, and the vectorization & argument-recycling of paste over vectors of strings to paste-together.

Answer 21

Vyxal, 8 bytes

ż\'vṀ\+j

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No clue why I didn't vectorise insertion back in May 2022.

Explained

żƛ?n\'Ṁ;\+j
żƛ           # For each item P in the range 1...len(input):
  ?n         #   at position P in the input,
    \'Ṁ      #   insert a '
       ;     # end map
        \+j  # join the result of that on "+" and output

Answer 22

Retina 0.8.2, 14 bytes

.
$`$&'$'+
.$

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.
$`$&'$'+

Replace each variable with its derivative multiplied by the other terms plus a trailing + to sum the products together.

.$

Delete the final trailing +.

Answer 23

Jelly, 10 bytes

ṛ€j”+sLj”'

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Port of DLosc's Pip solution.

 €            For each element of the input,
ṛ             the input.
  j”+         Join the copies on +,
     sL       split into slices the length of the original input,
       j”'    and join those on '.

Jelly, 11 bytes

p”'$ṛ¦ⱮJj”+

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The implementation of ¦ is a bit questionable, so we can just abuse Cartesian product to make up for it.

p”'            Pair each element of the input with ',
   $ṛ¦         and substitute elements of that into the input at
      ⱮJ       each index, individually.
        j”+    Join the results on +.

Just for fun, without preserving the order within each term, ṭ-Ƥ”'ż@j”+ is 10.

Answer 24

PARI/GP, 58 bytes

s->j=strjoin;j([j(strsplit(s,c),Str(c"'"))|c<-Vec(s)],"+")

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A port of @att's Mathematica answer.

Takes input as a string. Requires that all characters are unique.

---

PARI/GP, 64 bytes

v->i=1;j=strjoin;j([j([j(v[1..i]),j(v[i++..#v])],"'")|c<-v],"+")

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Take input as a list of characters.

Answer 25

05AB1E, 10 bytes

ε''«Nǝ}'+ý

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ε       map each character (of the implicit input string)
 ''     the ' character
 «      append to the current character
 N      iteration index
 ǝ      replace the input string's Nth character with the character with ' appended
}
'+      + character
ý       join by it

Answer 26

Raku, 33 bytes

This 33-byte version works on my local recent-ish version of Raku:

{join '+',.comb.map:{S/$^a/$a'/}}

But not on TIO, where a few more bytes are needed:

{join '+',.comb.map:{$^a;S/$a/$a'/}}

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• .comb breaks the input string up into characters.
• .map: { S/$^a/$a'/ } returns a copy of the input string for each character, in which that character has had an apostrophe appended to it.
• join '+' joins those strings together with a plus sign.

Answer 27

Japt, 10 bytes

¬£i''YÄÃq+

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¬£i''YÄÃq+     :Implicit input of string
¬              :Split
 £             :Map each element at 0-based index Y
  i''          :  Insert a "'" in U at index
     YÄ        :  Y+1
       Ã       :End map
        q+     :Join with "+"

Answer 28

Ruby, 36 bytes

->s{s.chars.map{s.sub _1,'\&\''}*?+}

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Answer 29

J-uby, 52 bytes

~(:-&(:& &:sub|(~:%&(~:+&?')|:& &:*)|:|&A)).D|~:*&?+

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Answer 30

Rust, 123 bytes

|s|{s.chars().map(|_|s).enumerate().map(|(i,w)|{let mut t=w.repeat(1);t.insert(i+1,'\'');t}).collect::<Vec<_>>().join("+")}

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