Sure! The sigmoid function, &#x03C3;(z)" role="presentation" style="position: relative;">σ(z)σ(z)\sigma(z), is defined as:

&#x03C3;(z)=11+e&#x2212;z" role="presentation" style="text-align: center; position: relative;">σ(z)=11+e−zσ(z)=11+e−z\sigma(z) = \frac{1}{1 + e^{-z}}

To find the partial derivative of &#x03C3;(z)" role="presentation" style="position: relative;">σ(z)σ(z)\sigma(z) with respect to z" role="presentation" style="position: relative;">zzz, we can use the chain rule:

&#x2202;&#x03C3;&#x2202;z=&#x2202;&#x2202;z(11+e&#x2212;z)" role="presentation" style="text-align: center; position: relative;">∂σ∂z=∂∂z(11+e−z)∂σ∂z=∂∂z(11+e−z)\frac{\partial \sigma}{\partial z} = \frac{\partial}{\partial z} \left( \frac{1}{1 + e^{-z}} \right)

=&#x2212;1(1+e&#x2212;z)2&#x2202;&#x2202;z(1+e&#x2212;z)" role="presentation" style="text-align: center; position: relative;">=−1(1+e−z)2∂∂z(1+e−z)=−1(1+e−z)2∂∂z(1+e−z)= -\frac{1}{(1 + e^{-z})^2} \frac{\partial}{\partial z} (1 + e^{-z})

=&#x2212;1(1+e&#x2212;z)2(&#x2212;e&#x2212;z)" role="presentation" style="text-align: center; position: relative;">=−1(1+e−z)2(−e−z)=−1(1+e−z)2(−e−z)= -\frac{1}{(1 + e^{-z})^2} (-e^{-z})

=e&#x2212;z(1+e&#x2212;z)2" role="presentation" style="text-align: center; position: relative;">=e−z(1+e−z)2=e−z(1+e−z)2= \frac{e^{-z}}{(1 + e^{-z})^2}

To simplify this expression, we can first note that:

&#x03C3;(z)=11+e&#x2212;z=ezez+1" role="presentation" style="text-align: center; position: relative;">σ(z)=11+e−z=ezez+1σ(z)=11+e−z=ezez+1\sigma(z) = \frac{1}{1 + e^{-z}} = \frac{e^z}{e^z + 1}

Then, we can rewrite the derivative using this form:

&#x2202;&#x03C3;&#x2202;z=e&#x2212;z(1+e&#x2212;z)2=1ez+1&#x22C5;ezez+1=&#x03C3;(z)&#x22C5;(1&#x2212;&#x03C3;(z))" role="presentation" style="text-align: center; position: relative;">∂σ∂z=e−z(1+e−z)2=1ez+1⋅ezez+1=σ(z)⋅(1−σ(z))∂σ∂z=e−z(1+e−z)2=1ez+1⋅ezez+1=σ(z)⋅(1−σ(z))\frac{\partial \sigma}{\partial z} = \frac{e^{-z}}{(1 + e^{-z})^2} = \frac{1}{e^{z}+1} \cdot \frac{e^{z}}{e^{z}+1} = \sigma(z) \cdot (1-\sigma(z))

So the partial derivative of the sigmoid function with respect to z" role="presentation" style="position: relative;">zzz can be expressed as a function of the sigmoid function itself: &#x03C3;(z)&#x22C5;(1&#x2212;&#x03C3;(z))" role="presentation" style="position: relative;">σ(z)⋅(1−σ(z))σ(z)⋅(1−σ(z))\sigma(z) \cdot (1-\sigma(z)).
물론이죠! 시그모이드 함수인 &#x03C3;(z)" role="presentation" style="위치: 상대적;">σ(z)σ(z)\sigma(z)는 다음과 같이 정의됩니다:&#x03C3;(z)=11+e&#x2212;z" role="presentation" style="text-align: 가운데; 위치: 상대;">σ(z)=11+e-zσ(z)=11+e-z\sigma(z) = \frac{1}{1 + e^{-z}}z에 대한 &#x03C3;(z)" role="presentation" style="위치: 상대적;">σ(z)σ(z)\sigma(z)의 편미분을 구하려면" role="presentation" style="위치: 상대적;">zzz, 연쇄 규칙을 사용할 수 있습니다:&#x2202;&#x03C3;&#x2202;z=&#x2202;&#x2202;z(11+e&#x2212;z)" role="presentation" style="text-align: 가운데; 위치: 상대;">∂σ∂z=∂∂z(11+e-z)∂σ∂z=∂∂z(11+e-z)\frac{\partial \sigma}{\partial z} = \frac{\partial}{\partial z} \왼쪽(\frac{1}{1 + e^{-z}} \right)=&#x2212;1(1+e&#x2212;z)2&#x2202;&#x2202;z(1+e&#x2212;z)" role="presentation" style="text-align: center; position: relative;">=-1(1+-e-z)2∂∂z(1+-e-z)=-1(1+-e-z)2∂∂z(1+-e-z)= -\frac{1}{(1 + e^{-z})^2} \frac{\partial}{\partial z} (1 + e^{-z})=&#x2212;1(1+e&#x2212;z)2(&#x2212;e&#x2212;z)" role="presentation" style="text-align: center; position: relative;">=-1(1+e-z)2(-e-z)=-1(1+e-z)2(-e-z)= -\frac{1}{(1 + e^{-z})^2} (-e^{-z})=e&#x2212;z(1+e&#x2212;z)2" role="presentation" style="text-align: 가운데; 위치: 상대;">=e-z(1+-e-z)2=e-z(1+-e-z)2= \frac{e^{-z}}{(1 + e^{-z})^2}이 식을 단순화하기 위해 먼저 주목할 수 있습니다:&#x03C3;(z)=11+e&#x2212;z=ezez+1" role="presentation" style="text-align: center; position: relative;">σ(z)=11+e-z=ezez+1σ(z)=11+e-z=ezez+1\sigma(z) = \frac{1}{1 + e^{-z}} = \frac{e^z}{e^z + 1}입니다.그런 다음 이 수식을 사용하여 도함수를 다시 작성할 수 있습니다:&#x2202;&#x03C3;&#x2202;z=e&#x2212;z(1+e&#x2212;z)2=1ez+1&#x22C5;ezez+1=&#x03C3;(z)&#x22C5;(1&#x2212;&#x03C3;(z))" role="presentation" style="text-align: center; position: 상대;">∂σ∂z=e-z(1+-e-z)2=1ez+1⋅ezez+1=σ(z)⋅(1-σ(z))∂σ∂z=e-z(1+-e-z)2=1ez+1⋅ezez+1=σ(z)⋅(1-σ(z))\frac{\partial \sigma}{\partial z} = \frac{e^{-z}}{(1 + e^{-z})^2} = \frac{1}{e^{-z}+1} \cdot \frac{e^{z}}{e^{z}+1} = \sigma(z) \cdot (1-\sigma(z))따라서 z에 대한 시그모이드 함수의 부분 도함수"role="presentation" style="위치: 상대;">zzz는 시그모이드 함수 자체의 함수로 표현할 수 있습니다: &#x03C3;(z)&#x22C5;(1&#x2212;&#x03C3;(z))" role="presentation" style="위치: 상대적;">σ(z)⋅(1-σ(z))σ(z)⋅(1-σ(z))\sigma(z)\cdot (1-\sigma(z)).