# sicp-ex-2.56

NTeGrotenhuis

Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule

d(u^n)/dr = nu^(n-1)(du/dr)

by adding a new clause to the deriv program and defining appropriate procedures exponentiation?, base, exponent, and make-exponentiation. (You may use the symbol ** to denote exponentiation.) Build in the rules that anything raised to the power 0 is 1 and anything raised to the power 1 is the thing itself.

code given in the book:

```  (define (deriv exp var)
(cond ((number? exp) 0)
((variable? exp)
(if (same-variable? exp var) 1 0))
((sum? exp)
(make-sum (deriv (addend exp) var)
(deriv (augend exp) var)))
((product? exp)
(make-sum
(make-product (multiplier exp)
(deriv (multiplicand exp) var))
(make-product (deriv (multiplier exp) var)
(multiplicand exp))))
(else
(error "unknown expression type -- DERIV" exp))))

(define (variable? x) (symbol? x))

(define (same-variable? v1 v2)
(and (variable? v1) (variable? v2) (eq? v1 v2)))

(define (make-sum a1 a2)
(cond ((=number? a1 0) a2)
((=number? a2 0) a1)
((and (number? a1) (number? a2)) (+ a1 a2))
(else (list '+ a1 a2))))

(define (make-product m1 m2)
(cond ((or (=number? m1 0) (=number? m2 0)) 0)
((=number? m1 1) m2)
((=number? m2 1) m1)
((and (number? m1) (number? m2)) (* m1 m2))
(else (list '* m1 m2))))

(define (=number? exp num)
(and (number? exp) (= exp num)))

(define (sum? x)
(and (pair? x) (eq? (car x) '+)))

(define (augend s) (caddr s))

(define (product? x)
(and (pair? x) (eq? (car x) '*)))

(define (multiplier p) (cadr p))

(define (multiplicand p) (caddr p))
```

First add this code to (derive exp var) to add differentiation of exponentiations.

``` ((exponentiation? exp)
(make-product
(make-product (exponent exp)
(make-exponentiation (base exp)
(make-sum(exponent exp) '-1)))
(deriv  (base exp) var)))
```

The end product is:

``` (define (deriv exp var)
(cond ((number? exp) 0)
((variable? exp)
(if (same-variable? exp var) 1 0))
((sum? exp)
(make-sum (deriv (addend exp) var)
(deriv (augend exp) var)))
((product? exp)
(make-sum
(make-product (multiplier exp)
(deriv (multiplicand exp) var))
(make-product (deriv (multiplier exp) var)
(multiplicand exp))))
((exponentiation? exp)
(make-product
(make-product (exponent exp)
(make-exponentiation (base exp)
(if (number? (exponent exp))
(- (exponent exp) 1)
(' (- (exponent exp) 1)))))
(deriv  (base exp) var)))
(else
(error "unknown expression type -- DERIV" exp))))
```

Next (define (exponentiation? exp))

``` (define (exponentiation? exp)
(and (pair? exp) (eq? (car exp) '**)))
```

We are using '** as the symbol for exponent.

Next, (define (base exp)) and (define (exponent exp))

```  (define (base exp)

(define (exponent exp)
```

All that is left is to (define (make-exponentiation base exp)).

``` (define (make-exponentiation base exp)
(cond ((=number? base 1) 1)
((=number? exp 1) base)
((=number? exp 0) 1)
(else
(list '** base exp))))
```

meteorgan

The above solution works. but in the end product, there is no need to check the exponent is number or not, procedure make-sum can do that. so the end product can be like this:

``` (define (deriv expr var)
(cond ((number? expr) 0)
((variable? expr)
(if (same-variable? expr var) 1 0))
((sum? expr)
(make-sum (deriv (addend expr) var)
(deriv (augend expr) var)))
((product? expr)
(make-sum
(make-product (multiplier expr)
(deriv (multiplicand expr) var))
(make-product (multiplicand expr)
(deriv (multiplier expr) var))))
((exponentiation? expr)
(make-product
(make-product
(exponent expr)
(make-exponentiation (base expr)
(make-sum (exponent expr) -1)))
(deriv (base expr) var)))
(else (error "unkown expression type -- DERIV" expr))))

```