What is the difference between (define (deriv-squared f ) (proc (deriv f))) and (define deriv-squared (proc deriv))?

The code in my homework works without error:

(define (deriv-squared f)
(square-a-procedure (deriv f)))

But when I define it as below, it doesn't evaluate and says: "deriv as undefined" even I define it. What is the difference between two definitions?

(define deriv-squared (square-a-procedure deriv))

Solution

A variable needs to have a binding before it it is referenced (used).

If you have

(define deriv-squared (square-a-procedure deriv))

then (square-a-procedure deriv) is evaluated and the resulting value is given the name deriv-squared. During the evaluation of (square-a-procedure deriv) the value of deriv is looked up. Therefore: the definition of deriv must be placed before the definition of deriv-squared.

(My guess is that you have placed the definition of deriv further down in the source file.)

Now why did your first definition work?

(define (deriv-squared f) (square-a-procedure (deriv f)))

This is short for:

(define deriv-squared 
   (lambda (f) 
      (square-a-procedure (deriv f))))

Here the (lambda (f) (square-a-procedure (deriv f)))) evaluated to a procedure. But deriv is not referenced until the procedure is used. So if deriv is defined before you use deriv-squared everything works fine.