Deep Learning · MT2 · section

Backpropagation with one hidden neuron

Neural-network foundations

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:

For the half squared loss used in this example:

$L=\frac12(\hat y-y)^2$ LaTeX: L=\frac12(\hat y-y)^2

Define a neuron's delta as the derivative of the loss with respect to its preactivation, not with respect to its output:

$\delta_o=\frac{\partial L}{\partial z_o}=(\hat y-y)g'(z_o)$ LaTeX: \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:

$\frac{\partial L}{\partial w_2}=\delta_o h,\qquad\frac{\partial L}{\partial b_2}=\delta_o$ LaTeX: \frac{\partial L}{\partial w_2}=\delta_o h,\qquad\frac{\partial L}{\partial b_2}=\delta_o

To reach the hidden neuron, continue backwards through the old output weight and the hidden activation:

$\delta_h=\delta_o w_2 f'(z_h)$ LaTeX: \delta_h=\delta_o w_2 f'(z_h)

$\frac{\partial L}{\partial w_1}=\delta_h x,\qquad\frac{\partial L}{\partial b_1}=\delta_h$ LaTeX: \frac{\partial L}{\partial w_1}=\delta_h x,\qquad\frac{\partial L}{\partial b_1}=\delta_h

The 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.

$z_h=W_1x+b_1$ LaTeX: z_h=W_1x+b_1

$h=f_1(z_h)$ LaTeX: h=f_1(z_h)

$z_o=W_2h+b_2$ LaTeX: z_o=W_2h+b_2

$o=f_2(z_o)$ LaTeX: o=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.

$L=\frac12\sum_j(o_j-y_j)^2$ LaTeX: L=\frac12\sum_j(o_j-y_j)^2

$\frac{\partial L}{\partial o_j}=o_j-y_j$ LaTeX: \frac{\partial L}{\partial o_j}=o_j-y_j

Yaad 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.

$\delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})$ LaTeX: \delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})

$\delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}$ LaTeX: \delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}

$\frac{\partial L}{\partial W_{ji}}=\delta_ja_i$ LaTeX: \frac{\partial L}{\partial W_{ji}}=\delta_ja_i

$\frac{\partial L}{\partial b_j}=\delta_j$ LaTeX: \frac{\partial L}{\partial b_j}=\delta_j

$W_{ji,\mathrm{new}}=W_{ji}-\eta\frac{\partial L}{\partial W_{ji}}$ LaTeX: W_{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

The forward pass evaluates hidden and output activations. The backward pass multiplies local derivatives through the same, unchanged weights. Tap the diagram to enlarge.

Step 1: forward propagation through the hidden neuron.

$z_h=0.5(2)+0.1=1.1$ LaTeX: z_h=0.5(2)+0.1=1.1

$h=\operatorname{ReLU}(1.1)=1.1$ LaTeX: h=\operatorname{ReLU}(1.1)=1.1

Step 2: forward propagation through the output neuron.

$z_o=0.8(1.1)+0.2=1.08$ LaTeX: z_o=0.8(1.1)+0.2=1.08

$\hat y=\operatorname{ReLU}(1.08)=1.08$ LaTeX: \hat y=\operatorname{ReLU}(1.08)=1.08

ReLU 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.

$L=\frac12(1.08-1)^2=0.0032$ LaTeX: L=\frac12(1.08-1)^2=0.0032

Both preactivations are positive, so both ReLU derivatives are 1.

$\delta_o=(1.08-1)(1)=0.08$ LaTeX: \delta_o=(1.08-1)(1)=0.08

Step 4: calculate the output-layer parameter gradients.

$\frac{\partial L}{\partial w_2}=0.08(1.1)=0.088$ LaTeX: \frac{\partial L}{\partial w_2}=0.08(1.1)=0.088

$\frac{\partial L}{\partial b_2}=0.08$ LaTeX: \frac{\partial L}{\partial b_2}=0.08

Step 5: propagate the derivative to the hidden neuron.

Use w₂ = 0.8, its value during the forward pass:

$\delta_h=0.08(0.8)(1)=0.064$ LaTeX: \delta_h=0.08(0.8)(1)=0.064

$\frac{\partial L}{\partial w_1}=0.064(2)=0.128$ LaTeX: \frac{\partial L}{\partial w_1}=0.064(2)=0.128

$\frac{\partial L}{\partial b_1}=0.064$ LaTeX: \frac{\partial L}{\partial b_1}=0.064

Step 6: update each parameter using its own gradient.

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.