<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>

CC BY-SA 4.0
2022-01-02T09:10:14.633Z
Could you please clarify: Is the variable $\omega$ a single weight connecting one input to one neuron in the hidden layer (layer $l-2$)?
Neil Slater
CC BY-SA 4.0
2022-01-02T10:04:49.153Z
yes, it's the weight connecting the first input layer to the first neurone in the layer layer l−2
Mylisteofanime nexv
<neural-networks><deep-learning><backpropagation><calculus>
Is my calculation of the partial derivative of the cost function with respect to a single weight in the first layer correct?
Mylisteofanime nexv
# 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  

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## 0 answers

