Newton's Method Help

For my Numerical Analysis class we are using Newton's Method to find the roots of a given function. The function given was "x = 2*sin(x)", and the answer we were given was "1.8954942670340", but my code returns -1.4014 after 7 iterations in the loop. For the variable "functn" I subtracted x in the orignal equation to get "2*sin(x) - x = 0". The tolerance we needed for the roots was 1e^-6. There is certainly an issue in the code, I just can't seem to find it. Thanks for reading, and looking this over.

clear all
close all
clc

functn = @(x) 2*sin(x) - x
interval = 0.001;  %small x-step used to approximate the function derivative
x = 0.01:0.01:1;
dydx = (functn(x+interval) - functn(x))/interval;
Root_value = -0.57;  % Function value
ytolerance = 1e-6;     % The convergence tolerance
initial_guess = 0.6;    % The initial guess for the location of the root x0

% set the starting function  value to the initial guess, and compute the initial error
x=initial_guess;
yerror = Root_value-functn(x);
check=1;
while abs(yerror)>ytolerance
    dydx = (functn(x+interval)-functn(x))/interval;
    dx = yerror/dydx;
    x=x+dx;
    disp("x"+check+" = "+x);
    disp(" ");
    yerror = Root_value - functn(x);
    check=check+1;
end

Answer 1

[David Hill]

f = @(x) 2*sin(x) - x;
fp= @(x) 2*cos(x) - 1;
tol=1e-10;
yerror=1;
g=1.2;%guess needs to be close enough to the root you are seeking
while abs(yerror)>tol
    yerror=f(g)/fp(g);
    g=g-yerror;
end

[David Hill]

If you estimate fp:

f = @(x) 2*sin(x) - x;
tol=1e-10;
yerror=1;
g=1.2;
while abs(yerror)>tol
    yerror=f(g)/((f(g+.001)-f(g))/.001);
    g=g-yerror;
end