Exercise 2.57.  Extend the differentiation program to handle sums and 
products of arbitrary numbers of (two or more) terms. Then the last 
example above could be expressed as

(deriv '(* x y (+ x 3)) 'x)

Try to do this by changing only the representation for sums and 
products, without changing the deriv procedure at all. For example, the 
addend of a sum would be the first term, and the augend would be the sum 
of the rest of the terms.

————————————————————————————————————————————————————————————————————————

sum? can stay as it is:

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

make-sum will take an arbitrary number of arguments:

(define (make-sum . as) (append '(+) as))

(define (addend s) (if (null? (cdr s)) 0 (cadr s)))

(define (augend s)
  (cond ((null? (cdr s)) 0)  ; augend of (+)
        ((null? (cddr s)) 0) ; augend of (+ x)
        ((null? (cdddr s)) (caddr s))
        (else (append '(+) (cddr s)))))

The changes for products would be analogous.