Newton's method for regression analysis without second derivative In regression analysis, instead of gradient descent, Newton's method can be used for minimizing the cost function.  However, in Newton's method, we need to calculate second derivative too.  
For example, to minimize a cost function $F(x)$, we need to find $x_0$ such that $F'(x0) = 0$, which means that we need to find the zeroes of $F'(x)$.
And for that, we can use Newton's method:
$$
x_1 = x_0 - \frac{F'(x_0)}{F"(x_0)}
$$
In the case of linear regression, the cost function is:
$$
        J = [h(x) - y]^2
$$
In the case of logistic regression, the cost function is:
$$
        J = y \log(h(x)) + (1 - y)(1 - \log(h(x)))
$$
In both the cases, since the cost function's minimum value is $0$, why can't we directly find the zeroes of the function using Newton's method, thus avoiding the calculation of the second derivative?  
 A: As mentioned in the comments, the reason is that the cost functions mentioned might not have any zeroes at all, in which case Newton's method will fail to find the minima.  
I have created a visualization to show this:  

As you can see, the method is not converging at all for this particular case.  
The code used to create this is stored here.  
Adding the relevant portion of the code here itself for convenience:  
newton.m:
% Dummy statement to avoid writing function in the first line and making it a 'function file' instead of a 'script file'
1;


% The function to find zeroes of.
% The function is specifically chosen to not have any zeroes
% so as to show the weakness of Newton's method.
function y = f(x)
    y = (x - 5).^2 + 5;
endfunction


% The derivative of f(x)
function y = fd(x)
    y = 2 * (x - 5);
endfunction


% Initial guess
x0 = 1.5;

% Max number of iterations
itermax = 20;

% Epsilon value initialized to a very large value
eps = 1;

% A vector for storing the history of the approximate roots
xvals = x0;

% Number of iterations done
itercount = 0;

% Required for plotting f(x) vs x
x = linspace(0, 10, 100);

% Create a figure whose output is not rendered on the screen
% Not working currently; supposedly a bug in Octave
% A workaround is to use gnuplot instead of qt - `graphics_toolkit gnuplot`
% but this is very slow.
% Uncomment the following to activate the feature once the bug is fixed
% figure('Visible','off');

% The main loop
while eps >= 1e-5 && itercount <= itermax
    % x1 = New value of root
    % x0 = Current value of root
    x1 = x0 - f(x0) / fd(x0);

    % Plot f(x)
    % Plot a line passing through points [x0, f(x0)] and [x1, 0]
    % Plot a line passing through points [x1, 0] and [x1, f(x1)]
    % Plot a line passing through points [x0, 0] and [x0, f(x0)]
    plot(x, f(x), ";f(x);", [x0 x1], [f(x0) 0], "-r;f'(x);", [x1 x1], [0 f(x1)], ":r", [x0 x0], [0 f(x0)], ":r");
    title('f(x) = (x-5)^2 + 5');


    % Set limits for the axes shown in the plots
    xlim([0 10]);
    ylim([0 30]);

    % Label the two consecutive zeroes on the X-axis
    text(x0, -2, sprintf('x%d', itercount), 'color', 'red');
    text(x1, -2, sprintf('x%d', itercount+1), 'color', 'red');

    % Print the plot to a file
    filename = sprintf('output/%05d.jpg', itercount);
    print(filename)

    % Append the zero to the array of zeroes calculated so far
    xvals = [xvals; x1];

    % Calculate the epsilon value
    eps = abs(x1-x0);

    x0 = x1;
    itercount = itercount+1;
end


% Print the result of the iteration
xvals
f_zero = f(xvals(end))
eps
itercount

