Deep Learning · MT2 · formulas
Layer forward pass
A layer does the neuron calculation for several receiving neurons. Finish all its outputs before using them as the next layer’s inputs. The bracketed superscript is a layer label, not a power.
z^{(l)}=W^{(l)}a^{(l-1)}+b^{(l)}a^{(l)}=f(z^{(l)})Forward propagation through multiple neurons
Definitions. A feedforward neural network has a directed acyclic computation graph: information moves from inputs toward outputs without recurrent feedback connections. A multilayer perceptron (MLP) is a feedforward network built from fully connected layers, with one or more hidden layers and nonlinear hidden activations.
Input, hidden and output layers
The input layer supplies the feature vector. A hidden layer learns an intermediate representation; its outputs are computed from preceding activations, not supplied as target labels. The output layer produces the prediction in the form needed by the task.
In a dense connection between two layers, every receiving neuron has a weight for every preceding activation. Each receiving neuron also normally has its own bias. An ANN is the broader family containing feedforward and recurrent models. A CNN can be feedforward without being an MLP, because convolution uses local shared connections.
General layer equations and dimensions
Let a⁽⁰⁾ = x. For layer l:
z^{(l)}=W^{(l)}a^{(l-1)}+b^{(l)}a^{(l)}=f^{(l)}(z^{(l)})If the preceding layer has n inputs and the current layer has m neurons, W has m rows and n columns. The vectors z, a and b have m components. Row j of W contains the incoming weights of neuron j.
W^{(l)}\in\mathbb R^{m\times n},\qquad b^{(l)},z^{(l)},a^{(l)}\in\mathbb R^{m\times1}For one hidden layer, these computations compose to give:
h=f^{(1)}(W^{(1)}x+b^{(1)})\hat y=f^{(2)}(W^{(2)}h+b^{(2)})The hidden activation must be nonlinear to obtain nonlinear representational power. Stacking affine layers alone still gives one affine mapping. A single threshold perceptron cannot separate XOR; a hidden nonlinear representation can make the two classes separable for the output layer.
Forward propagation versus training
Forward propagation evaluates these layer equations using the current parameters. Training additionally evaluates a loss, uses backpropagation to calculate gradients, and applies an optimizer update. The architecture specifies connections; backpropagation is a differentiation algorithm used to train that architecture.
Complete two-layer numerical
Question. A feedforward network receives x = [1, 0]ᵀ. Its hidden-layer weights are W¹ = LaTeX:
\begin{bmatrix}0.2 & 0.4\\0.3 & 0.7\end{bmatrix} and biases are b¹ = [0.1, 0.2]ᵀ. The output weights are W² = [0.5, 0.6] and output bias is b² = 0.1. Use ReLU in both layers and one row per receiving neuron. Calculate every preactivation and the final output.
Given
A feedforward network receives x = [1, 0]ᵀ.
Its hidden-layer weights are W¹ = LaTeX:
\begin{bmatrix}0.2 & 0.4\\0.3 & 0.7\end{bmatrix} and biases are b¹ = [0.1, 0.2]ᵀ.
The output weights are W² = [0.5, 0.6] and output bias is b² = 0.1.
Use ReLU in both layers and one row per receiving neuron.
To find
Find all preactivations, hidden activations and final outputs. Find the requested activation values and slopes.
Formula — which one and why
Forward propagation. Evaluate one layer completely before supplying its activated outputs to the next layer. Each layer has its own weights and biases.
x = input vector; h = hidden outputs; W₁,b₁ = hidden parameters; W₂,b₂ = output parameters; f₁,f₂ = the stated activations.
z_h=W_1x+b_1h=f_1(z_h)z_o=W_2h+b_2o=f_2(z_o)Yaad rakhne ka cue: Pehli layer ka activated output hi agli layer ka input hai.
ReLU and Leaky ReLU. ReLU returns zero for negative input and retains positive input. Leaky ReLU instead multiplies a negative input by its supplied slope.
z = preactivation; α (alpha) = supplied negative-side slope for Leaky ReLU. The derivative at zero needs an explicit convention.
\operatorname{ReLU}(z)=\max(0,z)\operatorname{LeakyReLU}(z)=\begin{cases}z&z\ge0\\\alpha z&z<0\end{cases}Yaad rakhne ka cue: Positive ko rehne do; negative ko zero. Leaky mein chhota negative rehne do.
Solution — step by step
Step 1: identify the matrix dimensions. W¹ is 2 × 2 and x is 2 × 1, so the hidden preactivation has two entries. W² is 1 × 2, so it takes those two activations and produces one output.
Step 2: calculate the first hidden preactivation. Multiply the first row by x, then add the first hidden bias.
z^{(1)}_1=0.2(1)+0.4(0)+0.1=0.3Step 3: calculate the second hidden preactivation.
z^{(1)}_2=0.3(1)+0.7(0)+0.2=0.5Step 4: activate the hidden layer. Both values are positive, so ReLU retains them.
a^{(1)}_1=\max(0,0.3)=0.3a^{(1)}_2=\max(0,0.5)=0.5a^{(1)}=\begin{bmatrix}0.3\\0.5\end{bmatrix}Step 5: calculate the output preactivation. Use the activated hidden values as this layer's inputs.
z^{(2)}=0.5(0.3)+0.6(0.5)+0.1z^{(2)}=0.15+0.30+0.10=0.55Step 6: apply the output activation.
\hat y=\max(0,0.55)=0.55Final answer: hidden preactivations and activations are [0.3, 0.5]ᵀ; output preactivation and prediction are both 0.55. They coincide here because every ReLU input is positive. No parameter has been updated during this forward pass.
Model answer: forward propagation, backpropagation and updating
Question
Question. Explain how a feedforward neural network learns using forward propagation, a loss function, backpropagation and gradient descent. State the order of operations.
Answer — step by step
Training repeatedly uses labeled examples to improve a network's weights and biases. The four operations have different jobs.
| Operation | Direction or action | Result |
|---|---|---|
| Forward propagation | Inputs move through successive layers | Predictions and intermediate activations |
| Loss calculation | Compare predictions with targets | A scalar objective for the example or batch |
| Backpropagation | Apply the chain rule from the loss toward earlier layers | Gradients of the objective with respect to parameters |
| Optimizer update | Use those gradients to change parameters | A new set of weights and biases |
1. Forward propagation
At each layer, calculate its weighted sums and activations. Retain these values because derivatives such as the sigmoid derivative depend on them. Complete a whole layer before calculating the next.
z^{(l)}=W^{(l)}a^{(l-1)}+b^{(l)},\qquad a^{(l)}=f(z^{(l)})2. Loss
Use the loss specified by the task. For one regression output, half squared loss is one possible choice:
L=\frac12(a-y)^2Here a is the prediction and y is the target. Other stated losses produce different derivatives.
3. Backpropagation
Multiply local derivatives along each dependency path. If a hidden unit influences several outputs, add the contributions from those branches. For a connection, the weight gradient is the receiving neuron's net-input derivative multiplied by the sending activation.
\frac{\partial L}{\partial w_{ji}}=\delta_j a_iHere δⱼ means the derivative of loss with respect to the receiving neuron's weighted sum. It is not automatically equal to raw prediction error; the activation and loss determine it.
4. Update and repeat
w_{\mathrm{new}}=w_{\mathrm{old}}-\eta\frac{\partial L}{\partial w}Compute all gradients using the same pre-update weights. Then update the requested parameters. Run another forward pass to calculate the new loss. Backpropagation calculates gradients; gradient descent uses them. Those are related, separate operations.
Backpropagation with one hidden neuron
Definition. Backpropagation computes the derivatives of a scalar objective with respect to network parameters by applying the chain rule in reverse through the computational graph. An optimizer then uses these derivatives to update the parameters.
Network equations and the meaning of delta
Consider one input, one hidden neuron and one output neuron:
z_h=w_1x+b_1,\qquad h=f(z_h)z_o=w_2h+b_2,\qquad\hat y=g(z_o)For the half squared loss used in this example:
L=\frac12(\hat y-y)^2Define a neuron's delta as the derivative of the loss with respect to its preactivation, not with respect to its output:
\delta_o=\frac{\partial L}{\partial z_o}=(\hat y-y)g'(z_o)The output-weight derivative contains the hidden activation because zₒ = w₂h + b₂:
\frac{\partial L}{\partial w_2}=\delta_o h,\qquad\frac{\partial L}{\partial b_2}=\delta_oTo reach the hidden neuron, continue backwards through the old output weight and the hidden activation:
\delta_h=\delta_o w_2 f'(z_h)\frac{\partial L}{\partial w_1}=\delta_h x,\qquad\frac{\partial L}{\partial b_1}=\delta_hThe bias derivative equals the corresponding delta because the derivative of z with respect to its additive bias is 1. All these derivatives belong to the same forward pass; an updated weight must not be used partway through this backward pass.
Complete worked numerical
Question. x = 2, y = 1; w₁ = 0.5, b₁ = 0.1; w₂ = 0.8, b₂ = 0.2. Both neurons use ReLU. Use half squared loss. Find the output and all four parameter gradients, then update all four parameters using η = 0.1.
Given
x = 2, y = 1.
w₁ = 0.5, b₁ = 0.1.
w₂ = 0.8, b₂ = 0.2.
Both neurons use ReLU.
Use half squared loss.
Find the output and all four parameter gradients, then update all four parameters using η = 0.1.
To find
Find all preactivations, hidden activations and final outputs. Calculate the half summed squared loss and its output derivatives. Find the requested deltas and parameter gradients, then update the specified parameters.
Formula — which one and why
Forward propagation. Evaluate one layer completely before supplying its activated outputs to the next layer. Each layer has its own weights and biases.
x = input vector; h = hidden outputs; W₁,b₁ = hidden parameters; W₂,b₂ = output parameters; f₁,f₂ = the stated activations.
z_h=W_1x+b_1h=f_1(z_h)z_o=W_2h+b_2o=f_2(z_o)Yaad rakhne ka cue: Pehli layer ka activated output hi agli layer ka input hai.
Half summed squared loss. This loss convention takes one-half of the sum of squared prediction errors. Use it when the question specifies it, or explicitly state it as an assumption when the paper omits the loss. Differentiating cancels the factor one-half against the square derivative.
oⱼ = output j; yⱼ = target j; L = half summed squared loss under the stated or explicitly assumed convention.
L=\frac12\sum_j(o_j-y_j)^2\frac{\partial L}{\partial o_j}=o_j-y_jYaad rakhne ka cue: Half aur square ka 2 cancel: output derivative = prediction − target.
Backpropagation. The loss tells us how each output must change. Multiply by that neuron’s activation slope to obtain its delta. A weight gradient is receiving delta × sending value. Hidden deltas must include every outgoing path.
δ (delta) = derivative of loss with respect to a neuron’s preactivation; f′ = activation derivative; aᵢ = sending activation; Wⱼᵢ = connection weight. Use the exact loss scaling in the question.
\delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})\delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}\frac{\partial L}{\partial W_{ji}}=\delta_ja_i\frac{\partial L}{\partial b_j}=\delta_jW_{ji,\mathrm{new}}=W_{ji}-\eta\frac{\partial L}{\partial W_{ji}}Yaad rakhne ka cue: Receiving delta × sending value = us connection ka gradient.
Solution — step by step
Step 1: forward propagation through the hidden neuron.
z_h=0.5(2)+0.1=1.1h=\operatorname{ReLU}(1.1)=1.1Step 2: forward propagation through the output neuron.
z_o=0.8(1.1)+0.2=1.08\hat y=\operatorname{ReLU}(1.08)=1.08ReLU is not bounded above by 1, so this output is valid. It is not being interpreted as a probability here.
Step 3: calculate the loss and output delta.
L=\frac12(1.08-1)^2=0.0032Both preactivations are positive, so both ReLU derivatives are 1.
\delta_o=(1.08-1)(1)=0.08Step 4: calculate the output-layer parameter gradients.
\frac{\partial L}{\partial w_2}=0.08(1.1)=0.088\frac{\partial L}{\partial b_2}=0.08Step 5: propagate the derivative to the hidden neuron.
Use w₂ = 0.8, its value during the forward pass:
\delta_h=0.08(0.8)(1)=0.064\frac{\partial L}{\partial w_1}=0.064(2)=0.128\frac{\partial L}{\partial b_1}=0.064Step 6: update each parameter using its own gradient.
w_{1,\mathrm{new}}=0.5-0.1(0.128)=0.4872b_{1,\mathrm{new}}=0.1-0.1(0.064)=0.0936w_{2,\mathrm{new}}=0.8-0.1(0.088)=0.7912b_{2,\mathrm{new}}=0.2-0.1(0.08)=0.192| Parameter | Gradient from the original forward pass | Updated value |
|---|---|---|
| w₁ | 0.128 | 0.4872 |
| b₁ | 0.064 | 0.0936 |
| w₂ | 0.088 | 0.7912 |
| b₂ | 0.08 | 0.192 |
Final answer: the original prediction is 1.08 and its loss is 0.0032. The table contains all requested gradients and updated parameters. A new forward pass would use all four new parameters together.
Drill 11 Full ReLU backpropagation
Question
Question. Input x = [1, 2]. Each row below lists the incoming weights of one receiving neuron. W₁ is shown below; b₁ = [0.1, 0.2]; W₂ is shown below; b₂ = [0.1]. Hidden and output activations are ReLU. Target y = [1]. Learning rate η = 0.1. Use half squared loss and update all weights and biases.
Answer — step by step
Given
Input x = [1, 2].
W₁ is printed with the question.
b₁ = [0.1, 0.2].
W₂ is printed with the question.
b₂ = [0.1].
Hidden and output activations are ReLU.
Target y = [1].
Learning rate η = 0.1.
Use half squared loss and update all weights and biases.
To find
Find all preactivations, hidden activations and final outputs. Calculate the half summed squared loss and its output derivatives. Find the requested deltas and parameter gradients, then update the specified parameters.
Formula — which one and why
Forward propagation. Evaluate one layer completely before supplying its activated outputs to the next layer. Each layer has its own weights and biases.
x = input vector; h = hidden outputs; W₁,b₁ = hidden parameters; W₂,b₂ = output parameters; f₁,f₂ = the stated activations.
z_h=W_1x+b_1h=f_1(z_h)z_o=W_2h+b_2o=f_2(z_o)Yaad rakhne ka cue: Pehli layer ka activated output hi agli layer ka input hai.
Half summed squared loss. This loss convention takes one-half of the sum of squared prediction errors. Use it when the question specifies it, or explicitly state it as an assumption when the paper omits the loss. Differentiating cancels the factor one-half against the square derivative.
oⱼ = output j; yⱼ = target j; L = half summed squared loss under the stated or explicitly assumed convention.
L=\frac12\sum_j(o_j-y_j)^2\frac{\partial L}{\partial o_j}=o_j-y_jYaad rakhne ka cue: Half aur square ka 2 cancel: output derivative = prediction − target.
Backpropagation. The loss tells us how each output must change. Multiply by that neuron’s activation slope to obtain its delta. A weight gradient is receiving delta × sending value. Hidden deltas must include every outgoing path.
δ (delta) = derivative of loss with respect to a neuron’s preactivation; f′ = activation derivative; aᵢ = sending activation; Wⱼᵢ = connection weight. Use the exact loss scaling in the question.
\delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})\delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}\frac{\partial L}{\partial W_{ji}}=\delta_ja_i\frac{\partial L}{\partial b_j}=\delta_jW_{ji,\mathrm{new}}=W_{ji}-\eta\frac{\partial L}{\partial W_{ji}}Yaad rakhne ka cue: Receiving delta × sending value = us connection ka gradient.
Solution — step by step
Formulas used before substituting numbers
For every receiving neuron, z is its weighted sum and a is its activated output. The half squared loss sums the errors from all output neurons.
z_j=\sum_iw_{ji}a_i+b_j,\quad a_j=f(z_j),\quad L=\frac12\sum_k(o_k-y_k)^2Delta means the loss gradient with respect to a neuron's raw score. First calculate output deltas, then hidden deltas using every connected output path.
\delta_{o_k}=(o_k-y_k)f'(z_{o_k})\delta_{h_j}=\left(\sum_kw_{kj}\delta_{o_k}\right)f'(z_{h_j})The gradient of a connection is its receiving delta multiplied by its sending activation. A bias receives the constant 1.
\frac{\partial L}{\partial w_{ji}}=\delta_j a_i,\qquad \frac{\partial L}{\partial b_j}=\delta_jw_{ji,\mathrm{new}}=w_{ji}-\eta\delta_j a_i,\quad b_{j,\mathrm{new}}=b_j-\eta\delta_jFor sigmoid, use f′ = a(1 − a). For ReLU, use f′ = 1 for a positive raw score and 0 for a negative raw score. Store the old weights until all gradients have been calculated.
Step 1 map the arrows. Each row of W₁ receives all inputs; each row of W₂ receives all hidden outputs. Bias is added once per receiving neuron. All gradients below use the OLD weights from this same forward pass.
Hidden neuron 1:
z_{h_1} = (0.2 \times 1) + (0.1 \times 2) + (0.1) = 0.5h_{1} = \max(0,0.5) = 0.5Hidden neuron 2:
z_{h_2} = (0.3 \times 1) + (-0.1 \times 2) + (0.2) = 0.3h_{2} = \max(0,0.3) = 0.3Output neuron 1:
z_{o_1} = (0.4 \times 0.5) + (0.5 \times 0.3) + (0.1) = 0.45o_{1} = \max(0,0.45) = 0.45Step 2 loss convention. This worked solution uses half summed squared loss, L = ½Σ(o − y)². Follow the loss specified in the question; where none is supplied, this is an explicit assumption.
Prediction minus target = [-0.55]. Squared errors = [0.3025]. Half-sum squared loss = 0.15125; mean squared error = 0.3025. These are different unless the number of output components makes the scale coincide.
Output delta 1:
\delta_{o_1} = 1 \times (0.45-1) \times 1 = -0.55The ReLU derivative is 1 here because this preactivation is positive.
Step 3 output weights. Gradient = receiving delta × sending activation. An output bias gradient equals its delta.
h1 → o1:
Subtract η × gradient = −0.0275. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies its activation.
g_{1,1}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,1}}=\delta_{o_1}h_1g_{1,1}^{(2)}=(-0.55)\times(0.5)\approx -0.275Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(2)}_{1,1,\mathrm{new}}=0.4-0.1(-0.275)\approx 0.4275h2 → o1:
Subtract η × gradient = −0.0165. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies its activation.
g_{1,2}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,2}}=\delta_{o_1}h_2g_{1,2}^{(2)}=(-0.55)\times(0.3)\approx -0.165Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(2)}_{1,2,\mathrm{new}}=0.5-0.1(-0.165)\approx 0.5165Step 4 hidden deltas. A hidden unit can influence EVERY connected output, so add those paths BEFORE multiplying by the hidden activation derivative.
Hidden 1 downstream contribution:
Calculate each outgoing path using the old outgoing weight and that output's delta.
Multiply within each path. Add the resulting contributions at the shared hidden neuron.
Contribution from output 1.
C_1=(0.4)\times(-0.55)\approx -0.22Accumulate the contributions, then apply the hidden activation derivative below.
\text{Sum}=C_1\approx -0.22\delta_{h_1} = -0.22 \times 1 = -0.22Hidden 2 downstream contribution:
Calculate each outgoing path using the old outgoing weight and that output's delta.
Multiply within each path. Add the resulting contributions at the shared hidden neuron.
Contribution from output 1.
C_1=(0.5)\times(-0.55)\approx -0.275Accumulate the contributions, then apply the hidden activation derivative below.
\text{Sum}=C_1\approx -0.275\delta_{h_2} = -0.275 \times 1 = -0.275Step 5 input weights. Multiply the hidden delta by the corresponding raw input, not by the hidden output.
x1 → h1:
Subtract η × gradient = −0.022. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{1,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,1}}=\delta_{h_1}x_1g_{1,1}^{(1)}=(-0.22)\times(1)\approx -0.22Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{1,1,\mathrm{new}}=0.2-0.1(-0.22)\approx 0.222x2 → h1:
Subtract η × gradient = −0.044. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{1,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,2}}=\delta_{h_1}x_2g_{1,2}^{(1)}=(-0.22)\times(2)\approx -0.44Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{1,2,\mathrm{new}}=0.1-0.1(-0.44)\approx 0.144x1 → h2:
Subtract η × gradient = −0.0275. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{2,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,1}}=\delta_{h_2}x_1g_{2,1}^{(1)}=(-0.275)\times(1)\approx -0.275Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{2,1,\mathrm{new}}=0.3-0.1(-0.275)\approx 0.3275x2 → h2:
Subtract η × gradient = −0.055. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{2,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,2}}=\delta_{h_2}x_2g_{2,2}^{(1)}=(-0.275)\times(2)\approx -0.55Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{2,2,\mathrm{new}}=-0.1-0.1(-0.55)\approx -0.045Step 6 bias updates. A bias receives a constant input 1, so its gradient equals the corresponding delta. Use the same old forward pass and learning rate as the weight updates.
b_{h_1,\mathrm{new}}=0.1-0.1(-0.220000000)=0.122000000b_{h_2,\mathrm{new}}=0.2-0.1(-0.275000000)=0.227500000b_{o_1,\mathrm{new}}=0.1-0.1(-0.550000000)=0.155000000These are independent per-neuron biases. If the question fixes biases, keep them fixed. If it specifies a single tied bias, add all deltas sharing that bias before updating it.
Drill 12 Two output sigmoid backpropagation
Question
Question. Input x = [0.2, 0.4]. Each row below lists the incoming weights of one receiving neuron. W₁ is shown below; b₁ = [0.1, 0.1]; W₂ is shown below; b₂ = [0, 0.1]. Hidden and output activations are sigmoid. Target y = [1, 0]. Learning rate η = 0.2. Use half summed squared loss and update all parameters.
Answer — step by step
Given
Input x = [0.2, 0.4].
W₁ is printed with the question.
b₁ = [0.1, 0.1].
W₂ is printed with the question.
b₂ = [0, 0.1].
Hidden and output activations are sigmoid.
Target y = [1, 0].
Learning rate η = 0.2.
Use half summed squared loss and update all parameters.
To find
Find all preactivations, hidden activations and final outputs. Calculate the half summed squared loss and its output derivatives. Find the sigmoid output and, if requested, its derivative. Find the requested deltas and parameter gradients, then update the specified parameters.
Formula — which one and why
Forward propagation. Evaluate one layer completely before supplying its activated outputs to the next layer. Each layer has its own weights and biases.
x = input vector; h = hidden outputs; W₁,b₁ = hidden parameters; W₂,b₂ = output parameters; f₁,f₂ = the stated activations.
z_h=W_1x+b_1h=f_1(z_h)z_o=W_2h+b_2o=f_2(z_o)Yaad rakhne ka cue: Pehli layer ka activated output hi agli layer ka input hai.
Half summed squared loss. This loss convention takes one-half of the sum of squared prediction errors. Use it when the question specifies it, or explicitly state it as an assumption when the paper omits the loss. Differentiating cancels the factor one-half against the square derivative.
oⱼ = output j; yⱼ = target j; L = half summed squared loss under the stated or explicitly assumed convention.
L=\frac12\sum_j(o_j-y_j)^2\frac{\partial L}{\partial o_j}=o_j-y_jYaad rakhne ka cue: Half aur square ka 2 cancel: output derivative = prediction − target.
Sigmoid and its slope. Use sigmoid when the neuron is specified to squash its preactivation into (0,1). Once its output is known, its derivative needs only output × (1−output).
z = preactivation; p = sigmoid output; e is the exponential base, not a prediction error in this formula.
p=\frac{1}{1+e^{-z}}\frac{dp}{dz}=p(1-p)Yaad rakhne ka cue: Sigmoid output p; uska slope p × (1−p).
Backpropagation. The loss tells us how each output must change. Multiply by that neuron’s activation slope to obtain its delta. A weight gradient is receiving delta × sending value. Hidden deltas must include every outgoing path.
δ (delta) = derivative of loss with respect to a neuron’s preactivation; f′ = activation derivative; aᵢ = sending activation; Wⱼᵢ = connection weight. Use the exact loss scaling in the question.
\delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})\delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}\frac{\partial L}{\partial W_{ji}}=\delta_ja_i\frac{\partial L}{\partial b_j}=\delta_jW_{ji,\mathrm{new}}=W_{ji}-\eta\frac{\partial L}{\partial W_{ji}}Yaad rakhne ka cue: Receiving delta × sending value = us connection ka gradient.
Solution — step by step
Formulas used before substituting numbers
For every receiving neuron, z is its weighted sum and a is its activated output. The half squared loss sums the errors from all output neurons.
z_j=\sum_iw_{ji}a_i+b_j,\quad a_j=f(z_j),\quad L=\frac12\sum_k(o_k-y_k)^2Delta means the loss gradient with respect to a neuron's raw score. First calculate output deltas, then hidden deltas using every connected output path.
\delta_{o_k}=(o_k-y_k)f'(z_{o_k})\delta_{h_j}=\left(\sum_kw_{kj}\delta_{o_k}\right)f'(z_{h_j})The gradient of a connection is its receiving delta multiplied by its sending activation. A bias receives the constant 1.
\frac{\partial L}{\partial w_{ji}}=\delta_j a_i,\qquad \frac{\partial L}{\partial b_j}=\delta_jw_{ji,\mathrm{new}}=w_{ji}-\eta\delta_j a_i,\quad b_{j,\mathrm{new}}=b_j-\eta\delta_jFor sigmoid, use f′ = a(1 − a). For ReLU, use f′ = 1 for a positive raw score and 0 for a negative raw score. Store the old weights until all gradients have been calculated.
Step 1 map the arrows. Each row of W₁ receives all inputs; each row of W₂ receives all hidden outputs. Bias is added once per receiving neuron. All gradients below use the OLD weights from this same forward pass.
Hidden neuron 1:
z_{h_1} = (0.1 \times 0.2) + (0.2 \times 0.4) + (0.1) = 0.2h_{1}=\frac{1}{1+e^{-(0.2)}}=0.549833997Hidden neuron 2:
z_{h_2} = (-0.2 \times 0.2) + (0.3 \times 0.4) + (0.1) = 0.18h_{2}=\frac{1}{1+e^{-(0.18)}}=0.544878892Output neuron 1:
z_{o_1} = (0.2 \times 0.549833997) + (-0.1 \times 0.544878892) + (0) = 0.05547891o_{1}=\frac{1}{1+e^{-(0.05547891)}}=0.513866171Output neuron 2:
z_{o_2} = (0.3 \times 0.549833997) + (0.4 \times 0.544878892) + (0.1) = 0.482901756o_{2}=\frac{1}{1+e^{-(0.482901756)}}=0.618432848Step 2 loss convention. This worked solution uses half summed squared loss, L = ½Σ(o − y)². Follow the loss specified in the question; where none is supplied, this is an explicit assumption.
Prediction minus target = [-0.486133829, 0.618432848]. Squared errors = [0.2363261, 0.382459187]. Half-sum squared loss = 0.309392643; mean squared error = 0.309392643. These are different unless the number of output components makes the scale coincide.
Output delta 1:
\delta_{o_1} = 1 \times (0.513866171-1) \times 0.249807729 = -0.121439988The derivative is o(1−o).
Output delta 2:
\delta_{o_2} = 1 \times (0.618432848-0) \times 0.235973661 = 0.145933863The derivative is o(1−o).
Step 3 output weights. Gradient = receiving delta × sending activation. An output bias gradient equals its delta.
h1 → o1:
Subtract η × gradient = −0.013354367. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies its activation.
g_{1,1}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,1}}=\delta_{o_1}h_1g_{1,1}^{(2)}=(-0.121439988)\times(0.549833997)\approx -0.066771834Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(2)}_{1,1,\mathrm{new}}=0.2-0.2(-0.066771834)\approx 0.213354367h2 → o1:
Subtract η × gradient = −0.013234017. This weight increases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies its activation.
g_{1,2}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,2}}=\delta_{o_1}h_2g_{1,2}^{(2)}=(-0.121439988)\times(0.544878892)\approx -0.066170086Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(2)}_{1,2,\mathrm{new}}=-0.1-0.2(-0.066170086)\approx -0.086765983h1 → o2:
Subtract η × gradient = 0.01604788. This weight decreases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies its activation.
g_{2,1}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{2,1}}=\delta_{o_2}h_1g_{2,1}^{(2)}=(0.145933863)\times(0.549833997)\approx 0.080239399Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(2)}_{2,1,\mathrm{new}}=0.3-0.2(0.080239399)\approx 0.28395212h2 → o2:
Subtract η × gradient = 0.015903256. This weight decreases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies its activation.
g_{2,2}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{2,2}}=\delta_{o_2}h_2g_{2,2}^{(2)}=(0.145933863)\times(0.544878892)\approx 0.079516282Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(2)}_{2,2,\mathrm{new}}=0.4-0.2(0.079516282)\approx 0.384096744Step 4 hidden deltas. A hidden unit can influence EVERY connected output, so add those paths BEFORE multiplying by the hidden activation derivative.
Hidden 1 downstream contribution:
Calculate each outgoing path using the old outgoing weight and that output's delta.
Multiply within each path. Add the resulting contributions at the shared hidden neuron.
Contribution from output 1.
C_1=(0.2)\times(-0.121439988)\approx -0.024287998Contribution from output 2.
C_2=(0.3)\times(0.145933863)\approx 0.043780159Accumulate the contributions, then apply the hidden activation derivative below.
\text{Sum}=C_1+C_2\approx 0.019492161\delta_{h_1} = 0.019492161 \times 0.247516573 = 0.004824633Hidden 2 downstream contribution:
Calculate each outgoing path using the old outgoing weight and that output's delta.
Multiply within each path. Add the resulting contributions at the shared hidden neuron.
Contribution from output 1.
C_1=(-0.1)\times(-0.121439988)\approx 0.012143999Contribution from output 2.
C_2=(0.4)\times(0.145933863)\approx 0.058373545Accumulate the contributions, then apply the hidden activation derivative below.
\text{Sum}=C_1+C_2\approx 0.070517544\delta_{h_2} = 0.070517544 \times 0.247985885 = 0.017487356Step 5 input weights. Multiply the hidden delta by the corresponding raw input, not by the hidden output.
x1 → h1:
Subtract η × gradient = 0.000192985. This weight decreases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{1,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,1}}=\delta_{h_1}x_1g_{1,1}^{(1)}=(0.004824633)\times(0.2)\approx 0.000964927Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{1,1,\mathrm{new}}=0.1-0.2(0.000964927)\approx 0.099807015x2 → h1:
Subtract η × gradient = 0.000385971. This weight decreases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{1,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,2}}=\delta_{h_1}x_2g_{1,2}^{(1)}=(0.004824633)\times(0.4)\approx 0.001929853Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{1,2,\mathrm{new}}=0.2-0.2(0.001929853)\approx 0.199614029x1 → h2:
Subtract η × gradient = 0.000699494. This weight decreases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{2,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,1}}=\delta_{h_2}x_1g_{2,1}^{(1)}=(0.017487356)\times(0.2)\approx 0.003497471Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{2,1,\mathrm{new}}=-0.2-0.2(0.003497471)\approx -0.200699494x2 → h2:
Subtract η × gradient = 0.001398988. This weight decreases.
The sending value and receiving delta have different roles. All deltas still use the original forward-pass weights.
The receiving neuron supplies the delta; the sending neuron supplies the input value.
g_{2,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,2}}=\delta_{h_2}x_2g_{2,2}^{(1)}=(0.017487356)\times(0.4)\approx 0.006994942Apply the learning rate to this gradient and subtract the result from the old weight.
W^{(1)}_{2,2,\mathrm{new}}=0.3-0.2(0.006994942)\approx 0.298601012Step 6 bias updates. A bias receives a constant input 1, so its gradient equals the corresponding delta. Use the same old forward pass and learning rate as the weight updates.
b_{h_1,\mathrm{new}}=0.1-0.2(0.004824633)=0.099035073b_{h_2,\mathrm{new}}=0.1-0.2(0.017487356)=0.096502529b_{o_1,\mathrm{new}}=0-0.2(-0.121439988)=0.024287998b_{o_2,\mathrm{new}}=0.1-0.2(0.145933863)=0.070813227These are independent per-neuron biases. If the question fixes biases, keep them fixed. If it specifies a single tied bias, add all deltas sharing that bias before updating it.