If I want to calculate derivatives at a point using finite difference method, how to set rhs in Matlab so it computes correctly?

Question

This question involves Matlab code, but is more about mathematics, so I post it here.

My code that calculates derivatives at a given point using finite difference method doesn't work and I think it's got something to do with rhs which I have given as rhs = [0;0;0;0;0;0] but previously I put it as rhs = [0;1;0;0;0;0] and it was even worse, because at least with all zeros I get the first dot correctly on the differentiated functions' graph.

Here is the picture that demostrates the problem:

The graph

function script()

I don't quite get what does rhs represent here and how to know the right values to it.

Here is my code:

syms x; f=tan(x^3) - x^2 + 1; h = 1;

xx=0:1/5:1;

yy = subs(f,xx);

ss=zeros(6,1);
rhs = [0;0;0;0;0;0];

l = 0;
r = 5;
 for j=0:l+r
        V(j+1,:)=[-l:r].^j;
    end
atr=V\rhs;

 for k=1:6
    ss(1)=ss(1)+atr(k)*yy(k)/h; 
 end

  l=1;r=4;   
    for j=0:l+r
        V(j+1,:)=[-l:r].^j;
    end
    atr=V\rhs; 

    for k=1:6
    ss(2)=ss(2)+atr(k)*yy(k)/h; 
    end

     l=2;r=3;

    for j=0:l+r
        V(j+1,:)=[-l:r].^j; 
    end
    atr=V\rhs; 

    for k=1:6
    ss(3)=ss(3)+atr(k)*yy(k)/h; 
    end

       l=3;r=2;

    for j=0:l+r
        V(j+1,:)=[-l:r].^j; 
    end
    atr=V\rhs; 


for k=1:6
    ss(4)=ss(4)+atr(k)*yy(k)/h;
    end

       l=4;r=1;

    for j=0:l+r
        V(j+1,:)=[-l:r].^j; 
    end
    atr=V\rhs; 

    for k=1:6
    ss(5)=ss(5)+atr(k)*yy(k)/h; 
    end

       l=5;r=0;

    for j=0:l+r
        V(j+1,:)=[-l:r].^j; 
    end
    atr=V\rhs; 

    for k=1:6
    ss(6)=ss(6)+atr(k)*yy(k)/h; 
    end

ss
subs(diff(f),xx)
    figure
    ezplot(diff(f),[-1, 1])
    hold on
    plot(xx,ss, '*')