Body: <p>I'm trying to understand the chain rule of backpropagation.
  This is what I understood. Is it correct?</p>
  <p><a href="https://i.sstatic.net/ttXSf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ttXSf.png" alt="enter image description here" /></a></p>
  <p><span class="math-container">$$ \frac{\partial E }{ \partial w} = \sum_{i} \frac{\partial E }{ \partial a_i^{(l)} } (\sum_{j} \frac{\partial a_i^{(l)} }{ \partial a_j^{(l-1)}  }(\sum_{k} \frac{\partial a_j^{(l-1)} }{ \partial a_k^{(l-2)}} \frac{\partial a_k^{(l-2)} }{ \partial z_k^{(l-2)}}  \frac{\partial z_k^{(l-2)} }{ \partial w}))$$</span></p>
  <ul>
  <li><span class="math-container">$ a_i^{(l)}$</span> is the activation of the neuron <span class="math-container">$i$</span> in the <span class="math-container">$l$</span>th layer</li>
  <li><span class="math-container">$ z_i^{(l)} $</span> is the sum of multiplication of weights and previous activations</li>
  <li><span class="math-container">$E$</span> is the error</li>
  <li><span class="math-container">$w$</span> is the weight</li>
  </ul>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-01-02T09:10:14.633Z
Comments.Text: Could you please clarify: Is the variable $\omega$ a single weight connecting one input to one neuron in the hidden layer (layer $l-2$)?
Comments.UserDisplayName: Neil Slater
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-01-02T10:04:49.153Z
Comments.Text: yes, it's the weight connecting the first input layer to the first neurone in the layer layer l−2
Comments.UserDisplayName: Mylisteofanime nexv
Tags: <neural-networks><deep-learning><backpropagation><calculus>
Title: Is my calculation of the partial derivative of the cost function with respect to a single weight in the first layer correct?
UserDisplayName: Mylisteofanime nexv
ThreadText: # Is my calculation of the partial derivative of the cost function with respect to a single weight in the first layer correct?
  
  (Asked by: username_0 on 2022-01-01)
  
  I'm trying to understand the chain rule of backpropagation.
  This is what I understood. Is it correct?
  
  [![enter image description here](https://i.sstatic.net/ttXSf.png)](https://i.sstatic.net/ttXSf.png)
  
  $$ \frac{\partial E }{ \partial w} = \sum\_{i} \frac{\partial E }{ \partial a\_i^{(l)} } (\sum\_{j} \frac{\partial a\_i^{(l)} }{ \partial a\_j^{(l-1)} }(\sum\_{k} \frac{\partial a\_j^{(l-1)} }{ \partial a\_k^{(l-2)}} \frac{\partial a\_k^{(l-2)} }{ \partial z\_k^{(l-2)}} \frac{\partial z\_k^{(l-2)} }{ \partial w}))$$
  
  * $ a\_i^{(l)}$ is the activation of the neuron $i$ in the $l$th layer
  * $ z\_i^{(l)} $ is the sum of multiplication of weights and previous activations
  * $E$ is the error
  * $w$ is the weight
  
  **Tags:** neural-networks, deep-learning, backpropagation, calculus
  > **Comments:**
  > **username_1**: Could you please clarify: Is the variable $\omega$ a single weight connecting one input to one neuron in the hidden layer (layer $l-2$)?  
  > **username_0**: yes, it's the weight connecting the first input layer to the first neurone in the layer layer l−2  
  
  ---
  
  ## 0 answers
  
  