A Step by Step Backpropagation Example


Backpropagation is a common method for training a neural network. There is no shortage of papers online that attempt to explain how backpropagation works, but few that include an example with actual numbers. This post is my attempt to explain how it works with a concrete example that folks can compare their own calculations to in order to ensure they understand backpropagation correctly.

Backpropagation in Python

You can play around with a Python script that I wrote that implements the backpropagation algorithm in this Github repo.

Backpropagation Visualization

For an interactive visualization showing a neural network as it learns, check out my Neural Network visualization.

Additional Resources

If you find this tutorial useful and want to continue learning about neural networks, machine learning, and deep learning, I highly recommend checking out Adrian Rosebrock’s new book, Deep Learning for Computer Vision with Python. I really enjoyed the book and will have a full review up soon.


For this tutorial, we’re going to use a neural network with two inputs, two hidden neurons, two output neurons. Additionally, the hidden and output neurons will include a bias.

Here’s the basic structure:

neural_network (7)

In order to have some numbers to work with, here are the initial weights, the biases, and training inputs/outputs:

neural_network (9)

The goal of backpropagation is to optimize the weights so that the neural network can learn how to correctly map arbitrary inputs to outputs.

For the rest of this tutorial we’re going to work with a single training set: given inputs 0.05 and 0.10, we want the neural network to output 0.01 and 0.99.

The Forward Pass

To begin, lets see what the neural network currently predicts given the weights and biases above and inputs of 0.05 and 0.10. To do this we’ll feed those inputs forward though the network.

We figure out the total net input to each hidden layer neuron, squash the total net input using an activation function (here we use the logistic function), then repeat the process with the output layer neurons.

Total net input is also referred to as just net input by some sources.

Here’s how we calculate the total net input for h_1:

net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1

net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775

We then squash it using the logistic function to get the output of h_1:

out_{h1} = \frac{1}{1+e^{-net_{h1}}} = \frac{1}{1+e^{-0.3775}} = 0.593269992

Carrying out the same process for h_2 we get:

out_{h2} = 0.596884378

We repeat this process for the output layer neurons, using the output from the hidden layer neurons as inputs.

Here’s the output for o_1:

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1

net_{o1} = 0.4 * 0.593269992 + 0.45 * 0.596884378 + 0.6 * 1 = 1.105905967

out_{o1} = \frac{1}{1+e^{-net_{o1}}} = \frac{1}{1+e^{-1.105905967}} = 0.75136507

And carrying out the same process for o_2 we get:

out_{o2} = 0.772928465

Calculating the Total Error

We can now calculate the error for each output neuron using the squared error function and sum them to get the total error:

E_{total} = \sum \frac{1}{2}(target - output)^{2}

Some sources refer to the target as the ideal and the output as the actual.
The \frac{1}{2} is included so that exponent is cancelled when we differentiate later on. The result is eventually multiplied by a learning rate anyway so it doesn’t matter that we introduce a constant here [1].

For example, the target output for o_1 is 0.01 but the neural network output 0.75136507, therefore its error is:

E_{o1} = \frac{1}{2}(target_{o1} - out_{o1})^{2} = \frac{1}{2}(0.01 - 0.75136507)^{2} = 0.274811083

Repeating this process for o_2 (remembering that the target is 0.99) we get:

E_{o2} = 0.023560026

The total error for the neural network is the sum of these errors:

E_{total} = E_{o1} + E_{o2} = 0.274811083 + 0.023560026 = 0.298371109

The Backwards Pass

Our goal with backpropagation is to update each of the weights in the network so that they cause the actual output to be closer the target output, thereby minimizing the error for each output neuron and the network as a whole.

Output Layer

Consider w_5. We want to know how much a change in w_5 affects the total error, aka \frac{\partial E_{total}}{\partial w_{5}}.

\frac{\partial E_{total}}{\partial w_{5}} is read as “the partial derivative of E_{total} with respect to w_{5}“. You can also say “the gradient with respect to w_{5}“.

By applying the chain rule we know that:

\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}

Visually, here’s what we’re doing:

output_1_backprop (4)

We need to figure out each piece in this equation.

First, how much does the total error change with respect to the output?

E_{total} = \frac{1}{2}(target_{o1} - out_{o1})^{2} + \frac{1}{2}(target_{o2} - out_{o2})^{2}

\frac{\partial E_{total}}{\partial out_{o1}} = 2 * \frac{1}{2}(target_{o1} - out_{o1})^{2 - 1} * -1 + 0

\frac{\partial E_{total}}{\partial out_{o1}} = -(target_{o1} - out_{o1}) = -(0.01 - 0.75136507) = 0.74136507

-(target - out) is sometimes expressed as out - target
When we take the partial derivative of the total error with respect to out_{o1}, the quantity \frac{1}{2}(target_{o2} - out_{o2})^{2} becomes zero because out_{o1} does not affect it which means we’re taking the derivative of a constant which is zero.

Next, how much does the output of o_1 change with respect to its total net input?

The partial derivative of the logistic function is the output multiplied by 1 minus the output:

out_{o1} = \frac{1}{1+e^{-net_{o1}}}

\frac{\partial out_{o1}}{\partial net_{o1}} = out_{o1}(1 - out_{o1}) = 0.75136507(1 - 0.75136507) = 0.186815602

Finally, how much does the total net input of o1 change with respect to w_5?

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1

\frac{\partial net_{o1}}{\partial w_{5}} = 1 * out_{h1} * w_5^{(1 - 1)} + 0 + 0 = out_{h1} = 0.593269992

Putting it all together:

\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}

\frac{\partial E_{total}}{\partial w_{5}} = 0.74136507 * 0.186815602 * 0.593269992 = 0.082167041

You’ll often see this calculation combined in the form of the delta rule:

\frac{\partial E_{total}}{\partial w_{5}} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1}) * out_{h1}

Alternatively, we have \frac{\partial E_{total}}{\partial out_{o1}} and \frac{\partial out_{o1}}{\partial net_{o1}} which can be written as \frac{\partial E_{total}}{\partial net_{o1}}, aka \delta_{o1} (the Greek letter delta) aka the node delta. We can use this to rewrite the calculation above:

\delta_{o1} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = \frac{\partial E_{total}}{\partial net_{o1}}

\delta_{o1} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1})


\frac{\partial E_{total}}{\partial w_{5}} = \delta_{o1} out_{h1}

Some sources extract the negative sign from \delta so it would be written as:

\frac{\partial E_{total}}{\partial w_{5}} = -\delta_{o1} out_{h1}

To decrease the error, we then subtract this value from the current weight (optionally multiplied by some learning rate, eta, which we’ll set to 0.5):

w_5^{+} = w_5 - \eta * \frac{\partial E_{total}}{\partial w_{5}} = 0.4 - 0.5 * 0.082167041 = 0.35891648

Some sources use \alpha (alpha) to represent the learning rate, others use \eta (eta), and others even use \epsilon (epsilon).

We can repeat this process to get the new weights w_6, w_7, and w_8:

w_6^{+} = 0.408666186

w_7^{+} = 0.511301270

w_8^{+} = 0.561370121

We perform the actual updates in the neural network after we have the new weights leading into the hidden layer neurons (ie, we use the original weights, not the updated weights, when we continue the backpropagation algorithm below).

Hidden Layer

Next, we’ll continue the backwards pass by calculating new values for w_1, w_2, w_3, and w_4.

Big picture, here’s what we need to figure out:

\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}



We’re going to use a similar process as we did for the output layer, but slightly different to account for the fact that the output of each hidden layer neuron contributes to the output (and therefore error) of multiple output neurons. We know that out_{h1} affects both out_{o1} and out_{o2} therefore the \frac{\partial E_{total}}{\partial out_{h1}} needs to take into consideration its effect on the both output neurons:

\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}}

Starting with \frac{\partial E_{o1}}{\partial out_{h1}}:

\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}}

We can calculate \frac{\partial E_{o1}}{\partial net_{o1}} using values we calculated earlier:

\frac{\partial E_{o1}}{\partial net_{o1}} = \frac{\partial E_{o1}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = 0.74136507 * 0.186815602 = 0.138498562

And \frac{\partial net_{o1}}{\partial out_{h1}} is equal to w_5:

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1

\frac{\partial net_{o1}}{\partial out_{h1}} = w_5 = 0.40

Plugging them in:

\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}} = 0.138498562 * 0.40 = 0.055399425

Following the same process for \frac{\partial E_{o2}}{\partial out_{h1}}, we get:

\frac{\partial E_{o2}}{\partial out_{h1}} = -0.019049119


\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}} = 0.055399425 + -0.019049119 = 0.036350306

Now that we have \frac{\partial E_{total}}{\partial out_{h1}}, we need to figure out \frac{\partial out_{h1}}{\partial net_{h1}} and then \frac{\partial net_{h1}}{\partial w} for each weight:

out_{h1} = \frac{1}{1+e^{-net_{h1}}}

\frac{\partial out_{h1}}{\partial net_{h1}} = out_{h1}(1 - out_{h1}) = 0.59326999(1 - 0.59326999 ) = 0.241300709

We calculate the partial derivative of the total net input to h_1 with respect to w_1 the same as we did for the output neuron:

net_{h1} = w_1 * i_1 + w_3 * i_2 + b_1 * 1

\frac{\partial net_{h1}}{\partial w_1} = i_1 = 0.05

Putting it all together:

\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}

\frac{\partial E_{total}}{\partial w_{1}} = 0.036350306 * 0.241300709 * 0.05 = 0.000438568

You might also see this written as:

\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\frac{\partial E_{total}}{\partial out_{o}} * \frac{\partial out_{o}}{\partial net_{o}} * \frac{\partial net_{o}}{\partial out_{h1}}}) * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}

\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\delta_{o} * w_{ho}}) * out_{h1}(1 - out_{h1}) * i_{1}

\frac{\partial E_{total}}{\partial w_{1}} = \delta_{h1}i_{1}

We can now update w_1:

w_1^{+} = w_1 - \eta * \frac{\partial E_{total}}{\partial w_{1}} = 0.15 - 0.5 * 0.000438568 = 0.149780716

Repeating this for w_2, w_3, and w_4

w_2^{+} = 0.19956143

w_3^{+} = 0.24975114

w_4^{+} = 0.29950229

Finally, we’ve updated all of our weights! When we fed forward the 0.05 and 0.1 inputs originally, the error on the network was 0.298371109. After this first round of backpropagation, the total error is now down to 0.291027924. It might not seem like much, but after repeating this process 10,000 times, for example, the error plummets to 0.0000351085. At this point, when we feed forward 0.05 and 0.1, the two outputs neurons generate 0.015912196 (vs 0.01 target) and 0.984065734 (vs 0.99 target).

If you’ve made it this far and found any errors in any of the above or can think of any ways to make it clearer for future readers, don’t hesitate to drop me a note. Thanks!

And while I have you…

In addition to dabbling in data science, I run Preceden timeline maker, the best timeline maker software on the web. If you ever need to create a high level timeline or roadmap to get organized or align your team, Preceden is a great option.

1,042 thoughts on “A Step by Step Backpropagation Example

  1. I think the formuale below is wrong. You should apply the multivariable version of the chaine rule instead.

    \frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}

  2. Hi, it seems your network visualization is giving an error. numInputs not defined. Please, let me know when it is working, I’d like to check whether I can reference or include it into my materials.
    Ingo Dahn

    • Although this is a good post. I don’t think the writer understand about bias, that’s why there’s no concrete explanation about how to update bias in this post.

      • The author likely assigned the same bias for the hidden and output neurons for simplicity. The article does not show the step of updating the bias weights, presumably as it is the exact same process as is used to update other weights (it is actually simpler, since the bias node activation is always 1.)

      • Correct observation. The bias is not updated, therefore the total error in the first round and the second round goes like this:
        Matt: E=0.298371109 -> E=0.291027924
        If you correctly update the bias also, then it should be like this:
        Barek: E=0.298371 -> E=0.280471
        When I did not update the bias, just like Matt, then I got
        Barek: E=0.298371 -> E=0.291028
        This proves that Matt did not update the bias.
        Anyway, this is a great post. Thank you, Matt!

  3. - xmioviettel.net

  4. hello!

    If there are multiple hidden layers, say two, three, or even more; How do you continue the back-propagation? Do you always take it two layers at a time?

    If we are dealing with the leftmost hidden layer, do we need to track how changes to weights in this layer effect the following multiple hidden layers and then finally the output layer (working out error)

    Tracing all routes that changes in a far left weight effect error seems to “blow up” very quickly so to speak.

    Can anyone point me towards a worked example with at least 1 more hidden layer?

  5. Machine Learning Roadmap 2022 - Codelivly

  6. Thank you for your detailed explanation.
    I wonder how you set the outputs to be 0.01 and 0.99 ?
    It is arbitrary chosen? Does the amount of the inputs and their vaules impact of the decesion how to set the outputs?

  7. This is multi-label classification, correct?
    Otherwise I would be wondering why we don’t either use a softmax in the last layer? Also our outputs don’t sum to 1.

    Thank you and kind regards

  8. How to calculate the total error from an arbitrary hidden layer in a neural network back propogation? - PhotoLens

  9. Hmm, in my case, after first round of backpropagation the total error is 0.291027774 instead 0.291027924. And after repeating 10,000 times, the total error is 0.0000351019 instead 0.0000351085. Тhe rest of the calculations matched perfectly

  10. Comprendre le DeepLearning et les réseaux de neurones en 10 minutes ! – TUTO3D

  11. 机器学习整理(神经网络) - 算法网

  12. I’m totally stumped, I’ve been over this article like 100 times and scrolled through as many comments and I cannot figure out how back propagation would work with multiple hidden layers. Using the example (I’m using d instead of the swirly thing because I’ve seen other comments do that and idk how to type the other character):

    dEtotal / dW1 = dEtotal / dOutputH1 * stuff I get
    dEtotal / dOutputH1 = dEo1/dNeto1 + same for O2
    dEo1/dNeto1 = dEo1 / dOutputo1 * dOutputo1 / dNeto1
    dEo1 / dOutputo1 = -(targeto1 – outputo1)

    I understand that for the output layer, but if you have a second hidden layer in place of the output layer then for some hn

    dEhn / dOutputhn = -(targethn – outputhn)
    What the hell is the target value? I’ve been pulling my hair out for like 4 hours and I cannot figure it out.

    • You need to work backwards through the layers and use the ‘delta’ from the next layer’s neurons. This delta should be stored whilst you are working backwards. The delta is the derivative of the current neurons output multiplied by the sum of all output neurons deltas * the weight between that neuron and the current one.

      So what you see in the blue box “You might also see this written as” works but replace input i, with the current neurons output passed through the derivative of the activation function

  13. Hi, i noticed that the value of out_h2 is incorrect. since its a logistic function, the sum of the two output can never exceed one but should approximate one in the nearest whole number. i go the value as 0.34468163

  14. OMG I’m a postgrad student and have learnt about CV,NLP which all covers NN but I’m still confusing with the math behind the learning. Thanks for your detailed formulas and calculations!!

  15. #2 BACKPROPAGATION algorithm. How a neural network learn ? A step by step demonstration. - DAOPOSTS

  16. I have written my own implementation and can verify I get the same results. Great! Now I am trying to add the bias.

    Firstly, My understanding was that the bias b1 would have 2 distinct weights being fed in to h1 and h2. And similarly b2 would have 2 distinct weights for o1 and o2. The article suggest that it is a single common weight for b1 -> h1,h2 and a single weight for b2 -> o1,o2. Which is correct?

    Secondly, when using the derivative of the activation function for the bias, is this just f'(1) always since the bias ‘output’ is always 1?

    • 1.
      You really should have distinct bias values for h1 and h2, but both can start from the given 0.35. These values will change during backpropagation. The same is true for o1 and o2. So after one backpropagation step:
      hidden_layer->b[0]=0.345614 hidden_layer->b[1]=0.345023
      output_layer->b[0]=0.530751 output_layer->b[1]=0.619049
      When using the derivative of the activation function, you start from out values and get back to net values. This is the “activation backward” step, and bias has no effect here. It will be updated in the next step, the “layer backward” step, which is matrix calculation.

  17. Answer Key 2022 Capf

  18. [Solved] My neural network backpropagates and converges at a total error of around 0.5 after many iterations. What have I done wrong? – Ten-tools.com

  19. Problema preciso con la función de activación sigmoide para Tensorflow/Keras 2.3.1 - neural-network en Español

  20. Critical Learning Periods in Deep Networks – Ramsey Elbasheer | History & ML

  21. Critical Learning Periods in Deep Networks – Commonwealth of Research & Analysis

  22. Machine Learning Roadmap 2022 – Codelivly

  23. Backpropagation from Perspektif Matematika dan Statistika - Techdigipro

  24. Very nicely explained. Can you please explain if there are more than one hidden layer what is done? I mean the repetitive portion which is iterated backward any worked example will be a great help.

  25. Classifying fruits with a Convolutional Neural Network in Keras | Daniel Pradilla

  26. Thank you very mach Matt.
    I develop my own Neural Net using Delphi, and the values are exactly the same!

    I adding hidden layers:
    1 Hidden layer -> 0.0000351020 (the same of the article)
    2 Hidden layer -> 0.0000061176
    4 Hidden layer -> 0.0000049050

    I put 0.5 to all bias(s), and change the learning rate (with 1 hidden layer):
    Learning rate 0.50 -> 0.0000277005
    Learning rate 0.75 -> 0.0000116993
    Learning rate 0.85 -> 0.0000086999

    Best Configuration (4 hidden layers, bias = 0.5, learning rate = 0.85)

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