(a) Write a context-free grammar for polynomials in x. Add semantic func- tions to produce an attribute grammar that will accumulate the polynomials derivative (as a string) in a synthesized attribute of the root of the parse tree. The following assumes that exponents in the input are all positive integers. PT more-Ts more. Ts. st :=T. d P. d:= more. Ts. d T num T tail > Ttail. c := num. v T. d: T. tail. d T tail x exp exp. c:- T. tail. c ► T. tail. d = exp. d T tail DT-tail. d:="" exp ** num exp. d := float_to_string(exp. cx num. v) + "* **" + int_to_string(num. - 1) expe exp. d := float_to_string(exp. c) more. Ts + + T more. Ts2 ► more. Ts2.st := more. TS. st + "+" +Td more. TS. d:= more. Ts2.d more. Ts € more-Ts. d:= more. Ts. st (b) Replace your semantic functions with action routines that can be evaluated during parsing. P + T {more. Ts. st := T. d} more_Ts {P. d := more-Ts. d} T + num {T-tail. c := num. v} T_tail {T. d := T-tail. d} T tail + x {exp. c := T-tail. c} exp {T-tail. d = exp. d} €{T-tail. d := ""} exp ** num {exp. d := float_to_string(exp. c x num. v) + "x **" + int. to. string(num. v - 1)} € {exp. d := float-to-string(exp. c)} more-Ts → + T {more-Ts2.st := more. Tsi. st + + T. d} more-T2 {more. Ts .d = more. Ts2.d} e{more. Ts. d := more. Ts. st} E Consider provided solutions of PLP Exercise 4.17 (1) Explain essential differences between solutions (a) and (b). (ii) Modify solutions (a), (b) so that the second derivative (not the first derivative) is accumulated in a synthetized attribute of the root of the parse tree.