# Gradient Descent in ReLU Neural Network

I’m new to machine learning and recently facing a problem on back propagation of training a neural network using ReLU activation function shown in the figure. My problem is to update the weights matrices in the hidden and output layers.

The cost function is given as:

$$J(\Theta) = \sum\limits_{i=1}^2 \frac{1}{2} \left(a_i^{(3)} - y_i\right)^2$$

where $$y_i$$ is the $$i$$-th output from output layer. Using the gradient descent algorithm, the weights matrices can be updated by:

$$\Theta_{jk}^{(2)} := \Theta_{jk}^{(2)} - \alpha\frac{\partial J(\Theta)}{\partial \Theta_{jk}^{(2)}}$$

$$\Theta_{ij}^{(3)} := \Theta_{ij}^{(3)} - \alpha\frac{\partial J(\Theta)}{\partial \Theta_{ij}^{(3)}}$$

I understand how to update the weight matrix at output layer $$\Theta_{ij}^{(3)}$$, however I don’t know how to update that from the input layer to hidden layer $$\Theta_{jk}^{(2)}$$ involving the ReLU activation units, i.e. not understanding how to get $$\frac{\partial J(\Theta)}{\partial \Theta_{jk}^{(2)}}$$.

Can anyone help me understand how to derive the gradient on the cost function...?

• if you found the solution .. can u please share it with us ? Nov 11, 2021 at 11:50

Have a look at this post. I found it quite useful when starting out with neural networks.

http://neuralnetworksanddeeplearning.com/chap2.html

• Thanks for your reference! That helps with a clearer picture of understanding, particularly for the calculus part. Nov 14, 2019 at 2:09

The derivative of a ReLU is:

$$\frac{\partial ReLU(x)}{\partial x} = \begin{cases} 0 & \text{if } x < 0 \\ 1 & \text{if } x > 0 \\ \end{cases}$$

So its value is set either to 0 or 1. It's not defined at 0, there must be a convention to set it either at 0 or 1 in this case.

To my understanding, it means that the error is either fully propagated to the previous layer (1), or completely stopped (0).

• I understand the special derivative for the ReLU function. But sorry for the unclear of my question that I want to understand the calculus part of the partial derivative, i.e. $\frac{\partial J(\Theta)}{\partial \Theta_{ij}^{(x)}}$, for some $x$-th layers $i$ and $j$. Thanks anyway! Nov 14, 2019 at 2:06
• maybe late, but this means that if we have a single final neuron that uses relu, if it predict 0, there is no "error propagation", since for backprop we always multiply by the derivative of the activation? Jun 30 at 16:45