# Calculate a derivative

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Compute and print a derivative of the symbolic expression a x^2 + b x + c in Mathematica syntax, preferably with simplification.

## Implementations

### C++

```#include <iostream>

using namespace std;

class Var;

class Base {
public:
virtual ~Base() {};
virtual const Base *clone() = 0;
virtual const Base *d(const string &v) const = 0;
virtual ostream &print(ostream &o) const = 0;
};

ostream &operator<<(ostream &o, const Base &e) { e.print(o); return o; }

class Int : public Base {
const int n;
public:
Int(int m) : n(m) {}
~Int() {}
const Base *clone() { return new Int(n); }
const Base *d(const string &v) const { return new Int(0); }
ostream &print(ostream &o) const { return o << n; }
};

class Var : public Base {
const string var;
public:
Var(string v) : var(v) {}
~Var() {}
const Base *clone() { return new Var(var); }
const Base *d(const string &v) const { return new Int(var == v ? 1 : 0); }
ostream &print(ostream &o) const { return o << var; }
};

class Plus : public Base {
const Base *e1, *e2;
public:
Plus(const Base *a, const Base *b) : e1(a), e2(b) {}
~Plus() { delete e1; delete e2; }
const Base *clone() { return new Plus(e1, e2); }
const Base *d(const string &v) const { return new Plus(e1->d(v), e2->d(v)); }
ostream &print(ostream &o) const
{ return o << "(" << *e1 << " + " << *e2 << ")"; }
};

class Times : public Base {
const Base *e1, *e2;
public:
Times(const Base *a, const Base *b) : e1(a), e2(b) {}
~Times() { delete e1; delete e2; }
const Base *clone() { return new Times(e1, e2); }
const Base *d(const string &v) const
{ return new Plus(new Times(e1, e2->d(v)), new Times(e1->d(v), e2)); }
ostream &print(ostream &o) const { return o << "(" << *e1 << " * " << *e2 << ")"; }
};

class Expr {
public:
Base *e;
Expr(Base *a) : e(a) {}
};

const Expr operator+(const Expr e1, const Expr e2)
{ return Expr(new Plus(e1.e->clone(), e2.e->clone())); }
const Expr operator*(const Expr e1, const Expr e2)
{ return Expr(new Times(e1.e->clone(), e2.e->clone())); }

ostream &operator<<(ostream &o, const Expr e) { return o << e.e; }

int main() {
Var vx("x"), va("a"), vb("b"), vc("c");
Expr x(&vx), a(&va), b(&vb), c(&vc);
Expr e = a*x*x + b*x + c;
cout << "d(" << *(e.e) << ", x) = " << *(e.e->d("x")) << endl;
return 0;
}
```

### Common Lisp

Lisp benefits from the use of the built-in s-expression type:

```(defun d (e x)
(cond
((atom e) (if (eq e x) 1 0))
((eq (car e) '+) `(+ ,(d (cadr e) x) ,(d (caddr e) x)))
((eq (car e) '*) `(+ (* ,(cadr e) ,(d (caddr e) x))
(* ,(caddr e) ,(d (cadr e) x))))))

(defvar *e* '(+ (+ (* a (* x x)) (* b x)) c))

(d *e* 'x)
```

The result is printed as:

`(+ (+ (+ (* A (+ (* X 1) (* X 1))) (* (* X X) 0)) (+ (* B 1) (* X 0))) 0)`

### F#

In F#, the + and * operators can be overloaded to work with the user-defined expr type:

```type expr =
| Int of int
| Add of expr * expr
| Mul of expr * expr
| Var of string with
static member ( + ) (f, g) = Add(f, g)
static member ( * ) (f, g) = Mul(f, g)
end

let rec d e x = match e with
| Int n -> Int 0
| Add(f, g) -> d f x + d g x
| Mul(f, g) -> f * d g x + g * d f x
| Var v -> Int (if v=x then 1 else 0)

let rec string_of = function
| Int n -> string_of_int n
| Var v -> v
| Add(f, g) -> "("^string_of f^" + "^string_of g^")"
| Mul(f, g) -> "("^string_of f^" * "^string_of g^")"

let x = Var "x" and a = Var "a" and b = Var "b" and c = Var "c"

let e = a * x * x + b * x + c

let () = Printf.printf "d %s x = %s\n" (string_of e) (string_of (d e "x"))
```

### Java

```abstract class Expr {
public abstract Expr d(String v);
public Expr plus(Expr e2) { return new Plus(this, e2); }
public Expr times(Expr e2) { return new Times(this, e2); }
}

class Int extends Expr {
private final int n;
public Int(int m) { n = m; }
public Expr d(String v) { return new Int(0); }
public String toString() { return Integer.toString(n); }
}

class Var extends Expr {
private final String var;
public Var(String v) { var = v; }
public Expr d(String v) { return new Int(var.equals(v) ? 1 : 0); }
public String toString() { return var; }
}

class Plus extends Expr {
private final Expr e1, e2;
public Plus(Expr a, Expr b) { e1 = a; e2 = b; }
public Expr d(String v) { return new Plus(e1.d(v), e2.d(v)); }
public String toString() { return "(" + e1 + " + " + e2 + ")"; }
}

class Times extends Expr {
private final Expr e1, e2;
public Times(Expr a, Expr b) { e1 = a; e2 = b; }
public Expr d(String v)
{ return new Plus(new Times(e1, e2.d(v)), new Times(e1.d(v), e2)); }
public String toString() { return "(" + e1 + " * " + e2 + ")"; }
}

public class Derivative {
public static void main(String[] args) {
Expr x = new Var("x"), a = new Var("a"),
b = new Var("b"), c = new Var("c");
Expr e = a.times(x).times(x).plus(b.times(x)).plus(c);
System.out.println("d(" + e + ", x) = " + e.d("x"));
}
}
```

### OCaml

Note that this implementation performs simple simplifications as expressions are constructed and pretty prints expressions with minimal bracketing (only sums inside products are bracketed).

```type expr =
| Int of int
| Var of string
| Add of expr * expr
| Mul of expr * expr

let ( +: ) f g = match f, g with
| Int n, Int m -> Int (n + m)
| Int 0, f
| f, Int 0 -> f
| f, g -> Add(f, g)

let ( *: ) f g = match f, g with
| Int n, Int m -> Int (n * m)
| Int 0, _
| _, Int 0 -> Int 0
| Int 1, f
| f, Int 1 -> f
| f, g -> Mul (f, g)

let rec d e x = match e with
| Int _ -> Int 0
| Add (f, g) -> d f x +: d g x
| Mul (f, g) -> f *: d g x +: g *: d f x
| Var v -> Int (if v=x then 1 else 0)

open Printf

let rec print ff = function
| Int n -> fprintf ff "%d" n
| Var v -> fprintf ff "%s" v
| Add (f, g) -> fprintf ff "%a + %a" print f print g
| Mul (f, g) ->
let product ff = function
| Add _ as expr -> fprintf ff "(%a)" print expr
| expr -> print ff expr
in
fprintf ff "%a %a" product f product g

let x = Var "x"
and a = Var "a"
and b = Var "b"
and c = Var "c"

let e = a *: x *: x +: b *: x +: c

let () = Printf.printf "d (%a) x = %a\n" print e print (d e "x")
```

The result is printed as:

`d (a x x + b x + c) x = a x + x a + b`

### Perl

```use Math::Symbolic qw();
my \$tree = Math::Symbolic->parse_from_string('a x^2 + b x + c');
\$tree->new(
'partial_derivative', \$tree, 'x'
)->apply_derivatives->simplify; # returns string
```

### Python

```class Expr:
return Plus(self, other)
def __mul__(self, other):
return Times(self, other)

class Int(Expr):
def __init__(self, n):
self.n = n
def d(self, v):
return Int(0)
def __str__(self):
return `self.n`

class Var(Expr):
def __init__(self, var):
self.var = var
def d(self, v):
return Int(self.var == v and 1 or 0)
def __str__(self):
return self.var

class Plus(Expr):
def __init__(self, a, b):
self.e1 = a
self.e2 = b
def d(self, v):
return Plus(self.e1.d(v), self.e2.d(v))
def __str__(self):
return "(%s + %s)" % (self.e1, self.e2)

class Times(Expr):
def __init__(self, a, b):
self.e1 = a
self.e2 = b
def d(self, v):
return Plus(Times(self.e1, self.e2.d(v)), Times(self.e1.d(v), self.e2))
def __str__(self):
return "(%s * %s)" % (self.e1, self.e2)

if __name__ == "__main__":
x = Var("x")
a = Var("a")
b = Var("b")
c = Var("c")
e = a * x * x + b * x + c
print "d(%s, x) = %s" % (e, e.d("x"))
```

### Ruby

```class Expr
def +(other) Plus.new(self, other) end
def *(other) Times.new(self, other) end
end

class Int < Expr
def initialize(n) @n = n end
def d(var) Int.new(0) end
def to_s() @n.to_s end
end

class Var < Expr
def initialize(var) @var = var end
def d(var) Int.new(@var==var ? 1 : 0) end
def to_s() @var end
end

class Plus < Expr
def initialize(a, b) @e1 = a; @e2 = b end
def d(var) Plus.new(@e1.d(var), @e2.d(var)) end
def to_s() "(#@e1 + #@e2)" end
end

class Times < Expr
def initialize(a, b) @e1 = a; @e2 = b end
def d(var) Plus.new(Times.new(@e1, @e2.d(var)), Times.new(@e1.d(var), @e2)) end
def to_s() "(#@e1 * #@e2)" end
end

x = Var.new("x")
a = Var.new("a")
b = Var.new("b")
c = Var.new("c")
e = a * x * x + b * x + c
puts "d(#{e}, x) = #{e.d("x")}"
```