<s>[INST] Interpolating $f$ and approximating its derivative We are given $x_0 = 1, x_1 = \frac{4}{3}$ and $x_2 = 2$.
Find a parabola which agrees with function $f(x) = (x + 1)\sin(x)$ in the given points. Afterwards derive a formula for the
approximation of the $f'$ i.e. the derivative in $x_1$. What is the approximation of $f'(x_1)$ for function f?
I did the first step using interpolation and got the polynomial $p(x) = -1.0647x^2 + 4.2390x - 1.4914.$ Now I do not know how to use this to approximate the derivative. Should I just compute the derivative of the polynomial $p$? What about the approximation then?
 [/INST] $$
f\left(x \right) = \left(x + 1\right)\sin x
\implies \left\lbrace
\begin{alignedat}{4}
  f\left(x_0\right)&= 2\sin \left(1\right) &&\approx 1.68,\qquad &\text{where}  \quad x_0&=1 \\
  f\left(x_1\right)&= \tfrac{7}{3}\sin \left(\tfrac{4}{3}\right)&\, &\approx 2.27, &\text{where}   \quad x_1&=4/3 \\
  f\left(x_2\right)&= 3\sin \left(2\right) &&\approx 2.73,& \text{where}   \quad x_2&=2& \\
\end{alignedat} \right.
$$
The general form of equation of parabola is 
$$
y-\widehat{y} = k\left(x-\widehat{x}\right)^2\qquad 
\text{ or }\qquad 
y\left(x\right) = ax^2+bx+c
$$
Substituting known values we get system of equations 
$$
\left\lbrace\begin{aligned} 
y\left(x_0\right) =f\left(x_0\right)\\ y\left(x_1\right) =f\left(x_1\right) \\ y\left(x_2\right) =f\left(x_2\right) 
\end{aligned}\right. \implies
\left\lbrace\begin{aligned}
 ax_0^2+bx_0+c &= 2\sin \left(1\right) \\
 ax_1^2+bx_1+c &= \tfrac{7}{3}\sin \left(\tfrac{4}{3}\right) \\
 ax_2^2+bx_2+c &= 3\sin \left(2\right) \\
\end{aligned}\right.
\iff
\left\lbrace\begin{aligned}
 a+b+c & \approx 1.68 \\
 \tfrac{16}{9}a+\tfrac{4}{3}b+c &\approx 2.27 \\
 4a +2b+c &\approx  2.73\\
\end{aligned}\right.
$$
Solving the last system of linear equations for coefficients $a,b$, and $c$ will give you general form of parabola.

As for approximating derivatives, I recommend following advise of @caverac and using Divided differences, or use straightforward Finite difference formula
$$
f'\left(x_1\right)\approx\frac{f\left(x_2\right)-f\left(x_0\right)}{x_2-x_0}
= \frac{2\sin 1 -3\sin 2}{1}\approx -1.045
$$
</s>