Exercise 2.56
Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule
d(x^n)/dx = n x^(n-1)
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.
这道题难度不大，先仿照 make-sum, make-product 写出 make-exponentiation 函数，剩下的部分就很简单了。
(define (make-exponentiation base exponent)
(cond ((=number? base 0) 0)
((=number? base 1) 1)
((and (number? base) (=number? exponent 0)) 1)
((and (number? base) (=number? exponent 1)) base)
((and (number? base) (number? exponent)) (* base (make-exponentiation base (- exponent 1))))
(else (list '** base exponent))))
(define (exponentiation? x) (and (pair? x) (eq? (car x) '**)))
(define (base e) (cadr e))
(define (exponent e) (caddr e))
(define (deriv exp var)
(cond ((number? exp) 0)
((variable? exp) (if (same-variable? exp var) 1 0))
((exponentiation? exp)
(let ((base (base exp))
(n (exponent exp)))
(if (same-variable? base var)
(make-product n
(make-exponentiation base (- n 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))))
(deriv '(** x 3) 'x)
(deriv '(** x 3) 'y)
; 执行结果
'(* 3 (** x 2))
0