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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _fD1(self, neg_pairs, A):
 """The gradient of the dissimilarity constraint function w.r.t. A. For example, let distance by L1 norm: f = f(\sum_{ij \in D} \sqrt{(x_i-x_j)A(x_i-x_j)'}) df/dA_{kl} = f'* d(\sum_{ij \in D} \sqrt{(x_i-x_j)^k*(x_i-x_j)^l})/dA_{kl} Note that d_ij*A*d_ij' = tr(d_ij*A*d_ij') = tr(d_ij'*d_ij*A) so, d(d_ij*A*d_ij')/dA = d_ij'*d_ij df/dA = f'(\sum_{ij \in D} \sqrt{tr(d_ij'*d_ij*A)}) * 0.5*(\sum_{ij \in D} (1/sqrt{tr(d_ij'*d_ij*A)})*(d_ij'*d_ij)) """

    dim = neg_pairs.shape[2]
    diff = neg_pairs[:, 0, :] - neg_pairs[:, 1, :]
    # outer products of all rows in `diff`
    M = np.einsum('ij,ik->ijk', diff, diff)
    # faster version of: dist = np.sqrt(np.sum(M * A[None,:,:], axis=(1,2)))
    dist = np.sqrt(np.einsum('ijk,jk', M, A))
    # faster version of: sum_deri = np.sum(M / (2 * (dist[:,None,None] + 1e-6)), axis=0)
    sum_deri = np.einsum('ijk,i->jk', M, 0.5 / (dist + 1e-6))
    sum_dist = dist.sum()
    return sum_deri / (sum_dist + 1e-6)