# Hidden delta, weight gradient and bias gradient

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/formulas/18

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/formulas/18/

A hidden neuron can influence several outputs. Add all those backward contributions before multiplying by its own activation derivative. For each connection, use the receiving neuron’s delta and the sending neuron’s activation. Keep the old weights unchanged until all gradients have been found.



$$
\delta_j^h=f\prime(z_j^h)\sum_k w_{kj}^{out}\delta_k^{out}
$$





$$
\frac{\partial L}{\partial w_{ji}}=\delta_j a_i
$$





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



## More outputs and vectorized backpropagation

**Definition.** Vectorized backpropagation expresses the chain rule using matrix products and elementwise derivatives. When a hidden activation affects several outputs, its derivative contains the sum of the contributions from every downstream path.

### Why the hidden derivative is a sum

If hidden activation hⱼ connects to output preactivations zₒ₁ and zₒ₂, changing hⱼ changes both outputs. The multivariable chain rule gives:



$$
\frac{\partial L}{\partial h_j}=W_{2,1j}\delta_{o_1}+W_{2,2j}\delta_{o_2}
$$



For any number of connected outputs, sum their contributions and then differentiate the hidden activation:



$$
\delta_{h_j}=\left(\sum_k W_{2,kj}\delta_{o_k}\right)f'(z_{h_j})
$$



Use the connecting weights from the original forward pass. Adding only the contribution of the first output would omit a dependency of the loss.

![The outgoing paths both contribute to the hidden derivative. Add the weighted output deltas, then apply the hidden activation derivative.](https://deep-learning-study-room.vercel.app/assets/figures/v_branch_gradient.svg)

Diagram: The outgoing paths both contribute to the hidden derivative. Add the weighted output deltas, then apply the hidden activation derivative.. Labels: Two output paths both affect the hidden neuron | Two output paths both affect the hidden neuron | Example: hidden activation h = 0.6, using sigmoid in the hidden neuron. | h = 0.6 | Output 1 delta = 0.2 | Connecting weight = 0.4 | Output 2 delta = −0.1 | Connecting weight = 0.5 | Downstream sum = 0.2 × 0.4 + (−0.1) × 0.5 = 0.03 | Hidden delta = 0.03 × 0.6 × (1 − 0.6) = 0.0072 | Add the two downstream paths first. Then multiply by the hidden activation | derivative.

The outgoing paths both contribute to the hidden derivative. Add the weighted output deltas, then apply the hidden activation derivative. Tap the diagram to enlarge.

### Matrix equations for one training sample

Use column vectors throughout. Let W₁ map d inputs to m hidden units and W₂ map those units to k outputs. The following output-delta formula assumes **half sum of squared errors** and an elementwise output activation f₂:



$$
L=\frac12\sum_{j=1}^{k}(\hat y_j-y_j)^2
$$





$$
\delta_2=(\hat y-y)\odot f_2'(z_2)
$$





$$
\frac{\partial L}{\partial W_2}=\delta_2a_1^T,\qquad\frac{\partial L}{\partial b_2}=\delta_2
$$





$$
\delta_1=(W_2^T\delta_2)\odot f_1'(z_1)
$$





$$
\frac{\partial L}{\partial W_1}=\delta_1x^T,\qquad\frac{\partial L}{\partial b_1}=\delta_1
$$



Here ⊙ means elementwise multiplication. The product δ₂a₁ᵀ is an **outer product**: every receiving delta is multiplied by every sending activation.

| Quantity | Shape | Reason |
| --- | --- | --- |
| W₁ and its gradient | m × d | m receiving hidden neurons, d inputs |
| W₂ and its gradient | k × m | k receiving output neurons, m hidden activations |
| δ₂ | k × 1 | One derivative per output preactivation |
| W₂ᵀδ₂ | m × 1 | The summed downstream derivative for each hidden unit |
| δ₂a₁ᵀ | (k × 1)(1 × m) = k × m | The output-weight gradient has the same shape as W₂ |

### How the loss changes the output delta

For MSE averaged over k output components, the squared-error derivative has the factor 2/k. For sigmoid with binary cross-entropy, or softmax with categorical cross-entropy and a normalized target, the paired derivative simplifies to prediction minus target:



$$
\delta_{\mathrm{output}}=\hat y-y
$$



That simplification already includes differentiation of the paired activation and loss. Multiplying by a sigmoid derivative again would be incorrect. For sigmoid with half squared error, the sigmoid derivative is still required.

For a batch, sum per-example parameter gradients; divide by the batch size only if the objective is a batch mean. State the output reduction and batch reduction separately when both appear in a question.

## 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:



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



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



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



The output-weight derivative contains the hidden activation because zₒ = w₂h + b₂:



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



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



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





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



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.



$$
z_h=W_1x+b_1
$$





$$
h=f_1(z_h)
$$





$$
z_o=W_2h+b_2
$$





$$
o=f_2(z_o)
$$



**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.



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





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



**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.



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





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





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





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





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



**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.](https://deep-learning-study-room.vercel.app/assets/figures/v_single_backprop.svg)

Diagram: The forward pass evaluates hidden and output activations. The backward pass multiplies local derivatives through the same, unchanged weights.. Labels: Follow this one-hidden-neuron example in both directions | Follow this one-hidden-neuron example in both directions | x = 2, target = 1. Both activations are ReLU; both raw sums are positive. | 2 | INPUT | 1.1 | HIDDEN | 1.08 | OUTPUT | weight 0.5 | weight 0.8 | 0.5 × 2 + 0.1 = 1.1 | 0.8 × 1.1 + 0.2 = 1.08 | Output delta = (1.08 − 1) × 1 = 0.08 | Back through old weight 0.8 | Hidden delta = 0.08 × 0.8 × 1 = 0.064 | For a weight gradient, multiply its receiving delta by the value entering that | connection.

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.**



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





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



**Step 2: forward propagation through the output neuron.**



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





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



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.**



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



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



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



**Step 4: calculate the output-layer parameter gradients.**



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





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



**Step 5: propagate the derivative to the hidden neuron.**

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



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





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





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



**Step 6: update each parameter using its own gradient.**



$$
w_{1,\mathrm{new}}=0.5-0.1(0.128)=0.4872
$$





$$
b_{1,\mathrm{new}}=0.1-0.1(0.064)=0.0936
$$





$$
w_{2,\mathrm{new}}=0.8-0.1(0.088)=0.7912
$$





$$
b_{2,\mathrm{new}}=0.2-0.1(0.08)=0.192
$$



| 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 13 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.

![Given matrices for this question](https://deep-learning-study-room.vercel.app/assets/figures/given_0601bfe757.svg)

Diagram: Given matrices for this question. Labels: GIVEN VALUES — read each matrix row from left to right | W₁ | 0.2 | 0.1 | 0.3 | −0.1 | W₂ | 0.4 | 0.5

Given matrices for this question Tap the diagram to enlarge.

![Worked example: all incoming weights and biases](https://deep-learning-study-room.vercel.app/assets/figures/net_drill13.svg)

Diagram: Worked example: all incoming weights and biases. Labels: Read each receiving neuron as one row of weights | INPUT | HIDDEN | OUTPUT | x1=1 | x2=2 | h1 | h2 | o1 | target 1 | h1 incoming: x1 × 0.2 + x2 × 0.1; bias 0.1 | h2 incoming: x1 × 0.3 + x2 × -0.1; bias 0.2 | o1 incoming: h1 × 0.4 + h2 × 0.5; bias 0.1 | Use these OLD connections for all gradients in the same pass.

Worked example: all incoming weights and biases Tap the diagram to enlarge.

## 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.



$$
z_h=W_1x+b_1
$$





$$
h=f_1(z_h)
$$





$$
z_o=W_2h+b_2
$$





$$
o=f_2(z_o)
$$



**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.



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





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



**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.



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





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





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





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





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



**Yaad rakhne ka cue:** Receiving delta × sending value = us connection ka gradient.

### Solution — step by step

### 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.



$$
z_j=\sum_iw_{ji}a_i+b_j,\quad a_j=f(z_j),\quad L=\frac12\sum_k(o_k-y_k)^2
$$



Delta 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.



$$
\frac{\partial L}{\partial w_{ji}}=\delta_j a_i,\qquad \frac{\partial L}{\partial b_j}=\delta_j
$$





$$
w_{ji,\mathrm{new}}=w_{ji}-\eta\delta_j a_i,\quad b_{j,\mathrm{new}}=b_j-\eta\delta_j
$$



For 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:**



$$
z_{h_1} = (0.2 \times  1) + (0.1 \times  2) + (0.1) = 0.5
$$





$$
h_{1} = \max(0,0.5) = 0.5
$$



**Hidden neuron 2:**



$$
z_{h_2} = (0.3 \times  1) + (-0.1 \times  2) + (0.2) = 0.3
$$





$$
h_{2} = \max(0,0.3) = 0.3
$$



**Output neuron 1:**



$$
z_{o_1} = (0.4 \times  0.5) + (0.5 \times  0.3) + (0.1) = 0.45
$$





$$
o_{1} = \max(0,0.45) = 0.45
$$



**Step 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:**



$$
\delta_{o_1} = 1 \times  (0.45-1) \times  1 = -0.55
$$



The 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:**

The calculation on h1 → o1

Sending activation**0.5**h1

×

Receiving delta**−0.55**δ at o1

=

Connection gradient**−0.275**h1 → o1

Now update this weight

Old weight**0.4**

→

New weight**0.4275**

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.



$$
g_{1,1}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,1}}=\delta_{o_1}h_1
$$





$$
g_{1,1}^{(2)}=(-0.55)\times(0.5)\approx -0.275
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(2)}_{1,1,\mathrm{new}}=0.4-0.1(-0.275)\approx 0.4275
$$



**h2 → o1:**

The calculation on h2 → o1

Sending activation**0.3**h2

×

Receiving delta**−0.55**δ at o1

=

Connection gradient**−0.165**h2 → o1

Now update this weight

Old weight**0.5**

→

New weight**0.5165**

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.



$$
g_{1,2}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,2}}=\delta_{o_1}h_2
$$





$$
g_{1,2}^{(2)}=(-0.55)\times(0.3)\approx -0.165
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(2)}_{1,2,\mathrm{new}}=0.5-0.1(-0.165)\approx 0.5165
$$



**Step 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.

Several output paths meet at this hidden neuron

Output 1

Output delta**−0.55**

×

Old outgoing weight**0.4**

=

Path contribution**−0.22**

↓

Add every path contribution**−0.22**Then multiply by the hidden activation derivative.

Multiply within each path. Add the resulting contributions at the shared hidden neuron.

**Contribution from output 1.**



$$
C_1=(0.4)\times(-0.55)\approx -0.22
$$



**Accumulate the contributions, then apply the hidden activation derivative below.**



$$
\text{Sum}=C_1\approx -0.22
$$





$$
\delta_{h_1} = -0.22 \times  1 = -0.22
$$



**Hidden 2 downstream contribution:**

Calculate each outgoing path using the **old** outgoing weight and that output's delta.

Several output paths meet at this hidden neuron

Output 1

Output delta**−0.55**

×

Old outgoing weight**0.5**

=

Path contribution**−0.275**

↓

Add every path contribution**−0.275**Then multiply by the hidden activation derivative.

Multiply within each path. Add the resulting contributions at the shared hidden neuron.

**Contribution from output 1.**



$$
C_1=(0.5)\times(-0.55)\approx -0.275
$$



**Accumulate the contributions, then apply the hidden activation derivative below.**



$$
\text{Sum}=C_1\approx -0.275
$$





$$
\delta_{h_2} = -0.275 \times  1 = -0.275
$$



**Step 5 input weights.** Multiply the hidden delta by the corresponding raw input, not by the hidden output.

**x1 → h1:**

The calculation on x1 → h1

Sending input**1**x1

×

Receiving delta**−0.22**δ at h1

=

Connection gradient**−0.22**x1 → h1

Now update this weight

Old weight**0.2**

→

New weight**0.222**

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.



$$
g_{1,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,1}}=\delta_{h_1}x_1
$$





$$
g_{1,1}^{(1)}=(-0.22)\times(1)\approx -0.22
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{1,1,\mathrm{new}}=0.2-0.1(-0.22)\approx 0.222
$$



**x2 → h1:**

The calculation on x2 → h1

Sending input**2**x2

×

Receiving delta**−0.22**δ at h1

=

Connection gradient**−0.44**x2 → h1

Now update this weight

Old weight**0.1**

→

New weight**0.144**

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.



$$
g_{1,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,2}}=\delta_{h_1}x_2
$$





$$
g_{1,2}^{(1)}=(-0.22)\times(2)\approx -0.44
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{1,2,\mathrm{new}}=0.1-0.1(-0.44)\approx 0.144
$$



**x1 → h2:**

The calculation on x1 → h2

Sending input**1**x1

×

Receiving delta**−0.275**δ at h2

=

Connection gradient**−0.275**x1 → h2

Now update this weight

Old weight**0.3**

→

New weight**0.3275**

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.



$$
g_{2,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,1}}=\delta_{h_2}x_1
$$





$$
g_{2,1}^{(1)}=(-0.275)\times(1)\approx -0.275
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{2,1,\mathrm{new}}=0.3-0.1(-0.275)\approx 0.3275
$$



**x2 → h2:**

The calculation on x2 → h2

Sending input**2**x2

×

Receiving delta**−0.275**δ at h2

=

Connection gradient**−0.55**x2 → h2

Now update this weight

Old weight**−0.1**

→

New weight**−0.045**

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.



$$
g_{2,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,2}}=\delta_{h_2}x_2
$$





$$
g_{2,2}^{(1)}=(-0.275)\times(2)\approx -0.55
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{2,2,\mathrm{new}}=-0.1-0.1(-0.55)\approx -0.045
$$



**Step 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.



$$
b_{h_1,\mathrm{new}}=0.1-0.1(-0.220000000)=0.122000000
$$





$$
b_{h_2,\mathrm{new}}=0.2-0.1(-0.275000000)=0.227500000
$$





$$
b_{o_1,\mathrm{new}}=0.1-0.1(-0.550000000)=0.155000000
$$



These 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 14 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.

![Given matrices for this question](https://deep-learning-study-room.vercel.app/assets/figures/given_907c03b5dc.svg)

Diagram: Given matrices for this question. Labels: GIVEN VALUES — read each matrix row from left to right | W₁ | 0.1 | 0.2 | −0.2 | 0.3 | W₂ | 0.2 | −0.1 | 0.3 | 0.4

Given matrices for this question Tap the diagram to enlarge.

![Worked example: both output paths must contribute to hidden gradients](https://deep-learning-study-room.vercel.app/assets/figures/net_drill14.svg)

Diagram: Worked example: both output paths must contribute to hidden gradients. Labels: Read each receiving neuron as one row of weights | INPUT | HIDDEN | OUTPUT | x1=0.2 | x2=0.4 | h1 | h2 | o1 | target 1 | o2 | target 0 | h1 incoming: x1 × 0.1 + x2 × 0.2; bias 0.1 | h2 incoming: x1 × -0.2 + x2 × 0.3; bias 0.1 | o1 incoming: h1 × 0.2 + h2 × -0.1; bias 0 | o2 incoming: h1 × 0.3 + h2 × 0.4; bias 0.1 | Use these OLD connections for all gradients in the same pass.

Worked example: both output paths must contribute to hidden gradients Tap the diagram to enlarge.

## 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.



$$
z_h=W_1x+b_1
$$





$$
h=f_1(z_h)
$$





$$
z_o=W_2h+b_2
$$





$$
o=f_2(z_o)
$$



**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.



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





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



**Yaad 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.



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





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





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





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





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



**Yaad rakhne ka cue:** Receiving delta × sending value = us connection ka gradient.

### Solution — step by step

### 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.



$$
z_j=\sum_iw_{ji}a_i+b_j,\quad a_j=f(z_j),\quad L=\frac12\sum_k(o_k-y_k)^2
$$



Delta 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.



$$
\frac{\partial L}{\partial w_{ji}}=\delta_j a_i,\qquad \frac{\partial L}{\partial b_j}=\delta_j
$$





$$
w_{ji,\mathrm{new}}=w_{ji}-\eta\delta_j a_i,\quad b_{j,\mathrm{new}}=b_j-\eta\delta_j
$$



For 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:**



$$
z_{h_1} = (0.1 \times  0.2) + (0.2 \times  0.4) + (0.1) = 0.2
$$





$$
h_{1}=\frac{1}{1+e^{-(0.2)}}=0.549833997
$$



**Hidden neuron 2:**



$$
z_{h_2} = (-0.2 \times  0.2) + (0.3 \times  0.4) + (0.1) = 0.18
$$





$$
h_{2}=\frac{1}{1+e^{-(0.18)}}=0.544878892
$$



**Output neuron 1:**



$$
z_{o_1} = (0.2 \times  0.549833997) + (-0.1 \times  0.544878892) + (0) = 0.05547891
$$





$$
o_{1}=\frac{1}{1+e^{-(0.05547891)}}=0.513866171
$$



**Output neuron 2:**



$$
z_{o_2} = (0.3 \times  0.549833997) + (0.4 \times  0.544878892) + (0.1) = 0.482901756
$$





$$
o_{2}=\frac{1}{1+e^{-(0.482901756)}}=0.618432848
$$



**Step 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:**



$$
\delta_{o_1} = 1 \times  (0.513866171-1) \times  0.249807729 = -0.121439988
$$



The derivative is o(1−o).

**Output delta 2:**



$$
\delta_{o_2} = 1 \times  (0.618432848-0) \times  0.235973661 = 0.145933863
$$



The derivative is o(1−o).

**Step 3 output weights.** Gradient = receiving delta × sending activation. An output bias gradient equals its delta.

**h1 → o1:**

The calculation on h1 → o1

Sending activation**0.549833997**h1

×

Receiving delta**−0.121439988**δ at o1

=

Connection gradient**−0.066771834**h1 → o1

Now update this weight

Old weight**0.2**

→

New weight**0.213354367**

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.



$$
g_{1,1}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,1}}=\delta_{o_1}h_1
$$





$$
g_{1,1}^{(2)}=(-0.121439988)\times(0.549833997)\approx -0.066771834
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(2)}_{1,1,\mathrm{new}}=0.2-0.2(-0.066771834)\approx 0.213354367
$$



**h2 → o1:**

The calculation on h2 → o1

Sending activation**0.544878892**h2

×

Receiving delta**−0.121439988**δ at o1

=

Connection gradient**−0.066170086**h2 → o1

Now update this weight

Old weight**−0.1**

→

New weight**−0.086765983**

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.



$$
g_{1,2}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{1,2}}=\delta_{o_1}h_2
$$





$$
g_{1,2}^{(2)}=(-0.121439988)\times(0.544878892)\approx -0.066170086
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(2)}_{1,2,\mathrm{new}}=-0.1-0.2(-0.066170086)\approx -0.086765983
$$



**h1 → o2:**

The calculation on h1 → o2

Sending activation**0.549833997**h1

×

Receiving delta**0.145933863**δ at o2

=

Connection gradient**0.080239399**h1 → o2

Now update this weight

Old weight**0.3**

→

New weight**0.28395212**

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.



$$
g_{2,1}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{2,1}}=\delta_{o_2}h_1
$$





$$
g_{2,1}^{(2)}=(0.145933863)\times(0.549833997)\approx 0.080239399
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(2)}_{2,1,\mathrm{new}}=0.3-0.2(0.080239399)\approx 0.28395212
$$



**h2 → o2:**

The calculation on h2 → o2

Sending activation**0.544878892**h2

×

Receiving delta**0.145933863**δ at o2

=

Connection gradient**0.079516282**h2 → o2

Now update this weight

Old weight**0.4**

→

New weight**0.384096744**

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.



$$
g_{2,2}^{(2)}=\frac{\partial L}{\partial W^{(2)}_{2,2}}=\delta_{o_2}h_2
$$





$$
g_{2,2}^{(2)}=(0.145933863)\times(0.544878892)\approx 0.079516282
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(2)}_{2,2,\mathrm{new}}=0.4-0.2(0.079516282)\approx 0.384096744
$$



**Step 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.

Several output paths meet at this hidden neuron

Output 1

Output delta**−0.121439988**

×

Old outgoing weight**0.2**

=

Path contribution**−0.024287998**

Output 2

Output delta**0.145933863**

×

Old outgoing weight**0.3**

=

Path contribution**0.043780159**

↓

Add every path contribution**0.019492161**Then multiply by the hidden activation derivative.

Multiply within each path. Add the resulting contributions at the shared hidden neuron.

**Contribution from output 1.**



$$
C_1=(0.2)\times(-0.121439988)\approx -0.024287998
$$



**Contribution from output 2.**



$$
C_2=(0.3)\times(0.145933863)\approx 0.043780159
$$



**Accumulate the contributions, then apply the hidden activation derivative below.**



$$
\text{Sum}=C_1+C_2\approx 0.019492161
$$





$$
\delta_{h_1} = 0.019492161 \times  0.247516573 = 0.004824633
$$



**Hidden 2 downstream contribution:**

Calculate each outgoing path using the **old** outgoing weight and that output's delta.

Several output paths meet at this hidden neuron

Output 1

Output delta**−0.121439988**

×

Old outgoing weight**−0.1**

=

Path contribution**0.012143999**

Output 2

Output delta**0.145933863**

×

Old outgoing weight**0.4**

=

Path contribution**0.058373545**

↓

Add every path contribution**0.070517544**Then multiply by the hidden activation derivative.

Multiply within each path. Add the resulting contributions at the shared hidden neuron.

**Contribution from output 1.**



$$
C_1=(-0.1)\times(-0.121439988)\approx 0.012143999
$$



**Contribution from output 2.**



$$
C_2=(0.4)\times(0.145933863)\approx 0.058373545
$$



**Accumulate the contributions, then apply the hidden activation derivative below.**



$$
\text{Sum}=C_1+C_2\approx 0.070517544
$$





$$
\delta_{h_2} = 0.070517544 \times  0.247985885 = 0.017487356
$$



**Step 5 input weights.** Multiply the hidden delta by the corresponding raw input, not by the hidden output.

**x1 → h1:**

The calculation on x1 → h1

Sending input**0.2**x1

×

Receiving delta**0.004824633**δ at h1

=

Connection gradient**0.000964927**x1 → h1

Now update this weight

Old weight**0.1**

→

New weight**0.099807015**

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.



$$
g_{1,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,1}}=\delta_{h_1}x_1
$$





$$
g_{1,1}^{(1)}=(0.004824633)\times(0.2)\approx 0.000964927
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{1,1,\mathrm{new}}=0.1-0.2(0.000964927)\approx 0.099807015
$$



**x2 → h1:**

The calculation on x2 → h1

Sending input**0.4**x2

×

Receiving delta**0.004824633**δ at h1

=

Connection gradient**0.001929853**x2 → h1

Now update this weight

Old weight**0.2**

→

New weight**0.199614029**

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.



$$
g_{1,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{1,2}}=\delta_{h_1}x_2
$$





$$
g_{1,2}^{(1)}=(0.004824633)\times(0.4)\approx 0.001929853
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{1,2,\mathrm{new}}=0.2-0.2(0.001929853)\approx 0.199614029
$$



**x1 → h2:**

The calculation on x1 → h2

Sending input**0.2**x1

×

Receiving delta**0.017487356**δ at h2

=

Connection gradient**0.003497471**x1 → h2

Now update this weight

Old weight**−0.2**

→

New weight**−0.200699494**

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.



$$
g_{2,1}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,1}}=\delta_{h_2}x_1
$$





$$
g_{2,1}^{(1)}=(0.017487356)\times(0.2)\approx 0.003497471
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{2,1,\mathrm{new}}=-0.2-0.2(0.003497471)\approx -0.200699494
$$



**x2 → h2:**

The calculation on x2 → h2

Sending input**0.4**x2

×

Receiving delta**0.017487356**δ at h2

=

Connection gradient**0.006994942**x2 → h2

Now update this weight

Old weight**0.3**

→

New weight**0.298601012**

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.



$$
g_{2,2}^{(1)}=\frac{\partial L}{\partial W^{(1)}_{2,2}}=\delta_{h_2}x_2
$$





$$
g_{2,2}^{(1)}=(0.017487356)\times(0.4)\approx 0.006994942
$$



Apply the learning rate to this gradient and subtract the result from the old weight.



$$
W^{(1)}_{2,2,\mathrm{new}}=0.3-0.2(0.006994942)\approx 0.298601012
$$



**Step 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.



$$
b_{h_1,\mathrm{new}}=0.1-0.2(0.004824633)=0.099035073
$$





$$
b_{h_2,\mathrm{new}}=0.1-0.2(0.017487356)=0.096502529
$$





$$
b_{o_1,\mathrm{new}}=0-0.2(-0.121439988)=0.024287998
$$





$$
b_{o_2,\mathrm{new}}=0.1-0.2(0.145933863)=0.070813227
$$



These 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 15 A negative ReLU blocks this path

## Question

**Question.** x = 2, w1 = −0.4, b1 = 0.1, w2 = 0.5, b2 = 0.2; hidden ReLU, linear output, y = 1, half squared loss. Find dw 1 and dw 2.

## Answer — step by step

### Given

x = 2; w₁ = −0.4; b₁ = 0.1; w₂ = 0.5; b₂ = 0.2.

Hidden activation: ReLU; output activation: linear.

Target y = 1; loss: half squared error.

### To find

Find the gradients of w₁ and w₂ and explain any zero gradients.

### 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.



$$
z_h=W_1x+b_1
$$





$$
h=f_1(z_h)
$$





$$
z_o=W_2h+b_2
$$





$$
o=f_2(z_o)
$$



**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.



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





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



**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.



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





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





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





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





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



**Yaad rakhne ka cue:** Receiving delta × sending value = us connection ka gradient.

### Solution — step by step

### Formulas used

The hidden neuron uses ReLU, but the output is linear. “Linear output” means its weighted sum is already its output.



$$
z_h=w_1x+b_1,\quad h=\max(0,z_h),\quad o=w_2h+b_2
$$





$$
L=\frac12(o-y)^2,\quad \delta_o=o-y
$$





$$
\frac{\partial L}{\partial w_2}=\delta_o h,\quad \delta_h=\delta_o w_2\operatorname{ReLU}'(z_h),\quad \frac{\partial L}{\partial w_1}=\delta_h x
$$



### Step 1: hidden weighted sum and activation



$$
z_h=(-0.4)\times2+0.1=-0.8+0.1=-0.7
$$





$$
h=\max(0,-0.7)=0
$$



Because the raw hidden score is negative, ReLU's derivative at this point is 0.

### Step 2: output prediction and output delta



$$
o=0.5\times0+0.2=0.2
$$





$$
\delta_o=0.2-1=-0.8
$$



### Step 3: output weight gradient



$$
\frac{\partial L}{\partial w_2}=(-0.8)\times0=0
$$



This is zero because the input to that connection, h, is zero.

### Step 4: hidden delta and input weight gradient



$$
\delta_h=(-0.8)\times0.5\times0=0
$$





$$
\frac{\partial L}{\partial w_1}=0\times2=0
$$



**Final answer:** both requested weight gradients are zero. The output bias gradient equals −0.8, so a zero weight gradient does not mean every parameter's gradient is zero.