Sure! The expression for z" role="presentation" style="position: relative;">zzz for an arbitrary neuron in the k" role="presentation" style="position: relative;">kkk-th layer can be written as:

zk=&#x2211;jwkjaj+bk z\_k = \sum\_{j} w\_{kj} a\_j + b\_k" role="presentation" style="text-align: center; position: relative;">zk=∑jwkjaj+bkzk=∑jwkjaj+bk z\_k = \sum\_{j} w\_{kj} a\_j + b\_k<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msub><mi>z</mi><mi>k</mi></msub><mo>=</mo><munder><mo>∑</mo><mi>j</mi></munder><msub><mi>w</mi><mrow><mi>k</mi><mi>j</mi></mrow></msub><msub><mi>a</mi><mi>j</mi></msub><mo>+</mo><msub><mi>b</mi><mi>k</mi></msub></mrow><annotation encoding="application/x-tex">z\_k = \sum\_{j} w\_{kj} a\_j + b\_k</annotation></semantics></math>zk​=j∑​wkj​aj​+bk​where wkj" role="presentation" style="position: relative;">wkjwkjw\_{kj} is the weight connecting neuron j" role="presentation" style="position: relative;">jjj in the (k&#x2212;1)" role="presentation" style="position: relative;">(k−1)(k−1)(k-1)-th layer to neuron k" role="presentation" style="position: relative;">kkk in the k" role="presentation" style="position: relative;">kkk-th layer, aj" role="presentation" style="position: relative;">ajaja\_j is the activation of neuron j" role="presentation" style="position: relative;">jjj in the (k&#x2212;1)" role="presentation" style="position: relative;">(k−1)(k−1)(k-1)-th layer, and bk" role="presentation" style="position: relative;">bkbkb\_k is the bias term for neuron k" role="presentation" style="position: relative;">kkk in the k" role="presentation" style="position: relative;">kkk-th layer.wkj" role="presentation" style="position: relative;">wkjwkjw\_{kj} is the weight connecting neuron j" role="presentation" style="position: relative;">jjj in the (k&#x2212;1)" role="presentation" style="position: relative;">(k−1)(k−1)(k-1)-th layer to neuron k" role="presentation" style="position: relative;">kkk in the k" role="presentation" style="position: relative;">kkk-th layer, aj" role="presentation" style="position: relative;">ajaja\_j is the activation of neuron j" role="presentation" style="position: relative;">jjj in the (k&#x2212;1)" role="presentation" style="position: relative;">(k−1)(k−1)(k-1)-th layer, and bk" role="presentation" style="position: relative;">bkbkb\_k is the bias term forwkj" role="presentation" style="position: relative;">wkjwkjw\_{kj} is the weight connecting neuron j" role="presentation" style="position: relative;">jjj in the (k&#x2212;1)" role="presentation" style="position: relative;">(k−1)(k−1)(k-1)-th layer to neuron k" role="presentation" style="position: relative;">kkk in the k" role="presentation" style="position: relative;">kkk-th layer, aj" role="presentation" style="position: relative;">ajaja\_j is the activation ofwkj" role="presentation" style="position: relative;">wkjwkjw\_{kj} is the weight connecting neuron $

The partial derivative of zk" role="presentation" style="position: relative;">zkzkz\_k with respect to the weight wkj" role="presentation" style="position: relative;">wkjwkjw\_{kj} can be found using the chain rule of differentiation as:zk" role="presentation" style="position: relative;">zkzkz\_k with respect to

&#x2202;zk&#x2202;wkj=aj \frac{\partial z\_k}{\partial w\_{kj}} = a\_j" role="presentation" style="text-align: center; position: relative;">∂zk∂wkj=aj∂zk∂wkj=aj \frac{\partial z\_k}{\partial w\_{kj}} = a\_j<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mi mathvariant="normal">∂</mi><msub><mi>z</mi><mi>k</mi></msub></mrow><mrow><mi mathvariant="normal">∂</mi><msub><mi>w</mi><mrow><mi>k</mi><mi>j</mi></mrow></msub></mrow></mfrac><mo>=</mo><msub><mi>a</mi><mi>j</mi></msub></mrow><annotation encoding="application/x-tex">\frac{\partial z\_k}{\partial w\_{kj}} = a\_j</annotation></semantics></math>∂wkj​∂zk​​=aj​The partial derivative of zk" role="presentation" style="position: relative;">zkzkz\_k with respect to the bias bk" role="presentation" style="position: relative;">bkbkb\_k can also be found using the chain rule of differentiation as:zk" role="presentation" style="position: relative;">zkzkz\_k with respect to the bias $

&#x2202;zk&#x2202;bk=1 \frac{\partial z\_k}{\partial b\_k} = 1" role="presentation" style="text-align: center; position: relative;">∂zk∂bk=1∂zk∂bk=1 \frac{\partial z\_k}{\partial b\_k} = 1<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mi mathvariant="normal">∂</mi><msub><mi>z</mi><mi>k</mi></msub></mrow><mrow><mi mathvariant="normal">∂</mi><msub><mi>b</mi><mi>k</mi></msub></mrow></mfrac><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\frac{\partial z\_k}{\partial b\_k} = 1</annotation></semantics></math>∂bk​∂zk​​=1I hope this helps!
물론이죠! k" role="presentation" style="위치: 상대적;">k" role="presentation" style="위치: 상대적;">kkk-th 레이어에 있는 임의의 뉴런에 대한 z의 표현식은 다음과 같이 작성할 수 있습니다:

zk=&#x2211;jwkjaj+bk z\_k = \sum\_{j} w\_{kj} a\_j + b\_k" role="presentation" style="text-align: 가운데; 위치: 상대적;">zk=∑jwkjaj+bkzk=∑jwkjaj+bk z\_k = \sum\_{j} w\_{kj} a\_j + b\_k<math xmlns="http://www. w3. org/1998/Math/MathML" display="block"><시맨틱><mrow><msub><mi>z</mi><mi>k</mi></msub><mo>=</mo><munder><mo>∑</mo><mi>j</mi></munder><msub><mi>w</mi><mrow><mi>k</mi><mi>j</mi></mrow></msub><msub><mi>a</mi><mi>j</mi></msub><mo>+</mo><msub><mi>b</mi><mi>k</mi></msub></mrow><annotation encoding="application/x- tex">z\_k = \sum\_{j} w\_{kj} a\_j + b\_k</annotation></semantics></math>zk=j∑wkjaj+bkwhere wkj" role="presentation" style="position: 상대적;">wkjwkjw\_{kj}는 뉴런 j를 연결하는 가중치입니다." role="presentation" style="위치: 상대적;">(k&#x2212;1)"에서 jjj" role="presentation" style="위치: 상대적;">(k-1)(k-1)(k-1)-세 번째 층에서 뉴런 k" role="presentation" style="위치: 상대적;">k" role="presentation" style="위치: 상대적;">kkk-세 번째 층, aj" role="presentation" style="위치: 상대적;">ajaja\_j는 뉴런 j의 활성화입니다." role="presentation" style="position: 상대적;">(k&#x2212;1)" role="presentation" style="위치: 상대적;">(k-1)(k-1)(k-1)-번째 층의 jjj이고, bk" role="presentation" style="위치: 상대적;">bkbkb\_k는 뉴런 k" role="presentation" style="위치: 상대적;">k" role="presentation" style="위치: 상대적;">k-th층에서 kk의 바이어스 용어입니다. wkj" role="presentation" style="위치: 상대적;">wkjwkjw\_{kj}는 뉴런 j를 연결하는 가중치입니다." role="presentation" style="위치: 상대적;">(k&#x2212;1)" role="presentation" style="position: 상대적;">(k-1)(k-1)(k-1)에서 뉴런 k까지" role="presentation" style="위치: 상대적;">kkk에서 k" role="presentation" style="위치: 상대적;">kkk-세 번째 층, aj" role="presentation" style="위치: 상대적;">ajaja\_j는 (k&#x2212;1)" role="presentation" style="위치: 상대적;">(k-1)(k-1)(k-1)-세 번째 층의 뉴런 j" role="presentation" style="위치: 상대적;">jjj, 그리고 bk" role="presentation" style="position: 상대적;">bkbkb\_k는 바이어스 항이고 wkj" role="presentation" style="위치: 상대적;">wkjwkjw\_{kj}는 뉴런 j를 연결하는 가중치입니다." role="presentation" style="위치: 상대적;">(k&#x2212;1)" role="presentation" style="위치: 상대적;">(k-1)(k-1)(k-1)-번째 층에서 뉴런 k" role="presentation" style="위치: 상대적;">k" role="presentation" style="위치: 상대적;">kkk-세 번째 레이어, aj" role="presentation" style="위치: 상대적;">ajaja\_j의 활성화입니다." role="presentation" style="위치: 상대적;">wkjwkjw\_{kj}는 뉴런을 연결하는 가중치 $입니다.

zk" role="presentation" style="위치: 상대적;">의 부분 미분은 가중치 wkj" role="presentation" style="위치: 상대적;">wkjwkjw\_{kj}에 대해 다음과 같이 미분 연쇄 규칙을 사용하여 구할 수 있습니다.

&#x2202;zk&#x2202;wkj=aj \frac{\partial z\_k}{\partial w\_{kj}} = a\_j" role="presentation" style="text-align 가운데; 위치: 상대적;">∂zk∂wkj=aj∂zk∂wkj=aj \frac{\partial z\_k}{\partial w\_{kj}} = a\_j<math xmlns="http://www. w3. org/1998/Math/MathML" display="block"><시맨틱스><mrow><mfrac><mrow><mi mathvariant="normal">∂</mi><msub><mi>z</mi><mi>k</mi></msub></mrow><mrow><mi mathvariant="normal">∂</mi><msub><mi>w</mi><mrow><mi>k</mi><mi>j</mi></mrow></msub></mrow></mfrac><mo>=</mo><msub><mi>a</mi><mi>j</mi></msub></mrow><annotation encoding="application/x-. tex">\frac{\부분 z\_k}{\부분 w\_{kj}} = a\_j</주석></semantics></math>∂wkj∂zk=ajzk의 부분 도함수" role="presentation" style="위치: 상대;">편향에 대한 zkzkz\_k bk" role="presentation" style="위치: 상대;">bkbkb\_k는 다음과 같이 미분 연쇄 규칙을 사용하여 찾을 수도 있습니다." role="presentation" style="위치: 상대;">편향에 대한 zkzkz\_k $.

&#x2202;zk&#x2202;bk=1 \frac{\partial z\_k}{\partial b\_k} = 1" role="presentation" style="text-align: 가운데; 위치: 상대적;">∂zk∂bk=1∂zk∂bk=1 \frac{\partial z\_k}{\partial b\_k} = 1<math xmlns="http://www.w3. org/1998/Math/MathML" display="block"><시맨틱스><mrow><mfrac><mrow><mi mathvariant="normal">∂</mi><msub><mi>z</mi><mi>k</mi></msub></mrow><mrow><mi mathvariant="normal">∂</mi><msub><mi>b</mi><mi>k</mi></msub></mrow></mfrac><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-. tex">\frac{\부분 z\_k}{\부분 b\_k} = 1</annotation></semantics></math>∂bk∂zk=1 도움이 되셨기를 바랍니다!