问如何在Octave中找到函数的导数？

EN

% this is just a formula to start with,  
% have fun and change it if you want to.  
f = @(x) x.^2 + 3*x - 1 + 5*x.*sin(x);  
% these next lines take the Anonymous function into a symbolic formula  
pkg load symbolic  
syms x;  
ff = f(x);  
% now calculate the derivative of the function  
ffd = diff(ff, x)  
% answer is ffd = (sym) 5*x*cos(x) + 2*x + 5*sin(x) + 3  
...

diff Difference and approximate derivative.
     diff(X), for a vector X, is [X(2)-X(1)  X(3)-X(2) ... X(n)-X(n-1)].
     diff(X), for a matrix X, is the matrix of row differences,
           [X(2:n,:) - X(1:n-1,:)].
     diff(X), for an N-D array X, is the difference along the first
          non-singleton dimension of X.
     diff(X,N) is the N-th order difference along the first non-singleton 
          dimension (denote it by DIM). If N >= size(X,DIM), diff takes 
          successive differences along the next non-singleton dimension.
     diff(X,N,DIM) is the Nth difference function along dimension DIM. 
         If N >= size(X,DIM), diff returns an empty array.

Examples:
   h = .001; x = 0:h:pi;
   diff(sin(x.^2))/h is an approximation to 2*cos(x.^2).*x
   diff((1:10).^2) is 3:2:19

   If X = [3 7 5
           0 9 2]
   then diff(X,1,1) is [-3 2 -3], diff(X,1,2) is [4 -2
                                                  9 -7],
   diff(X,2,2) is the 2nd order difference along the dimension 2, and
   diff(X,3,2) is the empty matrix.

xp= diff(xf);
yp= diff(yf);

% derivative:
dydx=yp./xp;
% also try:
dydx1=gradient(yf)./gradient(xf)

#This octave column vector is the result y axis/results of your given function  
#to which you want a derivative of.  Replace this vector with the results of  
#your function. 
observations = [2;8;3;4;5;9;10;5] 
 
#dy (aka the change in y) is the vertical distance (amount of change) between  
#each point and its prior.  The minus sign serves our purposes to "show me the  
#vertical different per unit x."  The NaN in square brackets does a 1 position 
#shift, since I want my tangent lines to be delimited by 1 step each. 
dy = observations - [NaN; observations(1:end-1,:)] 
 
#dx (aka the change in x) is the amount of horizontal distance (amount of  
#change) between each point and its prior.  for simplicity we make these all 1,  
#however your data points might not be constant width apart, that's okay  
#to populate the x vector here manually using the parameter-inputs to the  
#function that you want the derivative of. 
dx = ones(length(observations), 1) 
 
# -- alternative: if your x vector is subjective, then do something like: -- 
#function_parameters = [1;3;5;9;13;90;100;505] 
#dx = function_parameters 
 
#The derivative of your original function up top is "the slope of the  
#tangent line to the point on your curve.  The slope is calculated with the  
#equation: (rise / run) such that 'Rise' is y, and 'run' is x.  The tangent  
#line is calculated above with the dy variable.  'The point' is each point  
#in the observations, and 'the curve' is simply your function up top that  
#yielded those y results and x input parameters. 
dy_dx_dv1_macd = dy ./ dx 
 
#If this code scares you then this video goes into the eli5 detail about  
#what is going on and why every piece plays a role .  "The slope of a  
#tangent line to a point on your curve" is what a derivative is defined  
#to be.  And you can use octave (or any comptuer language) to calculate  
#it for your given function, and if you do this right and plot them,  
#double checking your work, you will find what your highschool teacher  
#was teaching by wrote: 'The derivative of x squared is x, and the  
#derivative of sin(x) is cos(x).'  It's beautiful masterpiece when you see  
#it fully because this is how physics around you isolated and captured  
#the information-gain to yield the nano machines you're using to read this sentence. 
# https://youtu.be/gtejJ3RCddE?t=5393