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.
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(check-equal? (deriv '(* x y) 'x) 'y) (check-equal? (deriv '(* x y (+ x 3)) 'x) '(+ (* x y) (* y (+ x 3))))