Exercise 2.56.  Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule

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.


(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)
(make-sum
(exponent exp)
-1)))
(deriv (base exp) var)))
(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 (addend s) (cadr s))
(define (augend s) (caddr s))
(define (product? x)
(and (pair? x) (eq? (car x) '*)))
(define (multiplier p) (cadr p))
(define (multiplicand p) (caddr p))
(define (base exp) (cadr exp))
(define (exponent exp) (caddr exp))
(define (make-exponentiation base exp)
(cond
((=number? base 1) 1)
((=number? exp 1) base)
((=number? exp 0) 1)
(else
(list '** base exp))))
(define (exponentiation? exp)
(and
(pair? exp)
(eq? (car exp) '**)))
(deriv (make-exponentiation 'x 3) 'x)

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