Write a similar python code :

clc; clear;

syms t$1$ t$2$ t$3$ t$4$ xbar $h$

syms u$1$ u$2$ u$3$ u$4$ u$($t$)$ dudt$($t$)$

$t1=x$ bar $-2^{**}h$

$t2=x$ bar $-h$;

$t3=$ xbar

$t4=$ xbar $+h$;

$\{$: $t2)(t4-t1)**(t4-t2)**(t4-t3)$

$\%$ Define right hand

$b=[u1$;$u2$;$u3$;$u4]$;

$\%$ solve for unknown coefficients

$c=\frac{A}{\frac{?}{?}}b$;

$$ne the polynomial in terms of Newton interpolation

$u(t)=c(1)+c(2)**(t-t1)+c(3)**(t-t1)**(t-t2)+c(4)**(t-t1)**(t-t2)**(t-t3)$;

$\%$ evaluate first derivative

dudt$(t)=$diff$(u,t)$;

$\frac{?}{?}\%$ evaluate $u\text{'}(xbar)$

disp$(u\text{'}(\frac{x}{b}ar)=)$

disp $(factor(dudt(xbar)))$

disp$(witht1=\frac{x}{b}ar-2*h$;$t2=\frac{x}{b}ar-h$;$t3=\frac{x}{b}ar$ ;$t4=\frac{x}{b}arth)$

disp$(andu1=u(t1)$;$u2=u(t2)$;$u3=u(t3)$;$u4=u(t4)$;$)$

$u^{\text{'}}(x$ bar $)=$

$[\frac{1}{6},u1-6**u2+3**u3+2**u4,\frac{1}{h}]$

with $t1=x$ bar $-{?}^{**}h$;$t2=x$ bar$-h$; $t3=$ xbar; $t4=$ xbarth

and $u1=u(t1)$;$u2=u(t2)$;$u3=u(t3)$;$u4=u(t4)$;