How to compute all second derivatives (only the diagonal of the Hessian matrix) in Tensorflow?

tf.gradients([f1,f2,f3],...) computes gradient of f=f1+f2+f3 Also, differentiating with respect to x[0] is problematic because x[0] refers to a new Slice node which is not an ancestor of your loss, so derivative with respect to it will be None. You could get around it by using pack to glue x[0], x[1], ... together into xx and have your loss depend on xx instead of x . An alternative is to use separate variables for individual components, in which case computing Hessian would look something like this.

def replace_none_with_zero(l):
  return [0 if i==None else i for i in l] 

tf.reset_default_graph()

x = tf.Variable(1.)
y = tf.Variable(1.)
loss = tf.square(x) + tf.square(y)
grads = tf.gradients([loss], [x, y])
hess0 = replace_none_with_zero(tf.gradients([grads[0]], [x, y]))
hess1 = replace_none_with_zero(tf.gradients([grads[1]], [x, y]))
hessian = tf.pack([tf.pack(hess0), tf.pack(hess1)])
sess = tf.InteractiveSession()
sess.run(tf.initialize_all_variables())
print hessian.eval()

You'll see

[[ 2.  0.]
 [ 0.  2.]]

The following function calculates 2nd derivatives (the diagonal of the Hessian matrix) in Tensorflow 2.0:

%tensorflow_version 2.x  # Tells Colab to load TF 2.x
import tensorflow as tf

def calc_hessian_diag(f, x):
    """
    Calculates the diagonal entries of the Hessian of the function f
    (which maps rank-1 tensors to scalars) at coordinates x (rank-1
    tensors).
    
    Let k be the number of points in x, and n be the dimensionality of
    each point. For each point k, the function returns

      (d^2f/dx_1^2, d^2f/dx_2^2, ..., d^2f/dx_n^2) .

    Inputs:
      f (function): Takes a shape-(k,n) tensor and outputs a
          shape-(k,) tensor.
      x (tf.Tensor): The points at which to evaluate the Laplacian
          of f. Shape = (k,n).
    
    Outputs:
      A tensor containing the diagonal entries of the Hessian of f at
      points x. Shape = (k,n).
    """
    # Use the unstacking and re-stacking trick, which comes
    # from https://github.com/xuzhiqin1990/laplacian/
    with tf.GradientTape(persistent=True) as g1:
        # Turn x into a list of n tensors of shape (k,)
        x_unstacked = tf.unstack(x, axis=1)
        g1.watch(x_unstacked)

        with tf.GradientTape() as g2:
            # Re-stack x before passing it into f
            x_stacked = tf.stack(x_unstacked, axis=1) # shape = (k,n)
            g2.watch(x_stacked)
            f_x = f(x_stacked) # shape = (k,)
        
        # Calculate gradient of f with respect to x
        df_dx = g2.gradient(f_x, x_stacked) # shape = (k,n)
        # Turn df/dx into a list of n tensors of shape (k,)
        df_dx_unstacked = tf.unstack(df_dx, axis=1)

    # Calculate 2nd derivatives
    d2f_dx2 = []
    for df_dxi,xi in zip(df_dx_unstacked, x_unstacked):
        # Take 2nd derivative of each dimension separately:
        #   d/dx_i (df/dx_i)
        d2f_dx2.append(g1.gradient(df_dxi, xi))
    
    # Stack 2nd derivates
    d2f_dx2_stacked = tf.stack(d2f_dx2, axis=1) # shape = (k,n)
    
    return d2f_dx2_stacked

Here's an example usage, with the function f(x) = ln(r), where x are 3D coordinates and r is the radius is spherical coordinates:

f = lambda q : tf.math.log(tf.math.reduce_sum(q**2, axis=1))
x = tf.random.uniform((5,3))

d2f_dx2 = calc_hessian_diag(f, x)
print(d2f_dx2)

The will look something like this:

tf.Tensor(
[[ 1.415968    1.0215727  -0.25363517]
 [-0.67299247  2.4847088   0.70901346]
 [ 1.9416015  -1.1799507   1.3937857 ]
 [ 1.4748447   0.59702784 -0.52290654]
 [ 1.1786096   0.07442689  0.2396735 ]], shape=(5, 3), dtype=float32)

We can check the correctness of the implementation by calculating the Laplacian (i.e., by summing the diagonal of the Hessian matrix), and comparing to the theoretical answer for our chosen function, 2 / r^2:

print(tf.reduce_sum(d2f_dx2, axis=1)) # Laplacian from summing above results
print(2./tf.math.reduce_sum(x**2, axis=1)) # Analytic expression for Lapalcian

I get the following:

tf.Tensor([2.1839054 2.5207298 2.1554365 1.5489659 1.49271  ], shape=(5,), dtype=float32)
tf.Tensor([2.1839058 2.5207298 2.1554365 1.5489662 1.4927098], shape=(5,), dtype=float32)

They agree to within rounding error.