title: Derivative of ReLU
body: <p>I'm learning PyTorch. Here is the first example in official tutorial. I got two questions, as shown in the block below, </p>
  
  <p>a) I understand that derivative of a ReLU function is 0 when x &lt; 0 and 1 when x > 0. Is that right? But the code seems to keep the x > 0 part unchanged and set x &lt; 0 part to 0. Why is that?</p>
  
  <p>b) Why transpose, i.e. x.T.mm(grad_h)? A transpose does't seem needed to me. I'm just confused. Thanks,</p>
  
  <pre><code># -*- coding: utf-8 -*-
  
  import torch
  
  
  dtype = torch.float
  device = torch.device("cpu")
  # device = torch.device("cuda:0") # Uncomment this to run on GPU
  
  # N is batch size; D_in is input dimension;
  # H is hidden dimension; D_out is output dimension.
  N, D_in, H, D_out = 64, 1000, 100, 10
  
  # Create random input and output data
  x = torch.randn(N, D_in, device=device, dtype=dtype)
  y = torch.randn(N, D_out, device=device, dtype=dtype)
  
  # Randomly initialize weights
  w1 = torch.randn(D_in, H, device=device, dtype=dtype)
  w2 = torch.randn(H, D_out, device=device, dtype=dtype)
  
  learning_rate = 1e-6
  for t in range(500):
      # Forward pass: compute predicted y
      h = x.mm(w1)
      h_relu = h.clamp(min=0)
      y_pred = h_relu.mm(w2)
  
      # Compute and print loss
      loss = (y_pred - y).pow(2).sum().item()
      print(t, loss)
  
      # Backprop to compute gradients of w1 and w2 with respect to loss
      grad_y_pred = 2.0 * (y_pred - y)
      grad_w2 = h_relu.t().mm(grad_y_pred)
      grad_h_relu = grad_y_pred.mm(w2.t())
  </code></pre>
  
  <blockquote>
  <pre><code>    grad_h = grad_h_relu.clone()
      grad_h[h &lt; 0] = 0
      grad_w1 = x.t().mm(grad_h)
  </code></pre>
  </blockquote>
  
  <pre><code>    # Update weights using gradient descent
      w1 -= learning_rate * grad_w1
      w2 -= learning_rate * grad_w2
  </code></pre>