Sure, the sigmoid function can be written as:

&#x3C3;(z)=11+e&#x2212;z \sigma(z) = \frac{1}{1 + e^{-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}}<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><msup><mi>e</mi><mrow><mo>−</mo><mi>z</mi></mrow></msup></mrow></mfrac></mrow><annotation encoding="application/x-tex">\sigma(z) = \frac{1}{1 + e^{-z}}</annotation></semantics></math>σ(z)=1+e−z1​where z" role="presentation" style="position: relative;">zzz is the weighted sum of the inputs plus the bias.

The partial derivative of the sigmoid function with respect to z" role="presentation" style="position: relative;">zzz can be found using the quotient rule of differentiation as:z" role="presentation" style="position: relative;">zzz can be found using the quotient rule of differentiation as

&#x2202;&#x3C3;&#x2202;z=&#x2202;&#x2202;z(11+e&#x2212;z)=e&#x2212;z(1+e&#x2212;z)2=&#x3C3;(z)(1&#x2212;&#x3C3;(z)) \begin{aligned}\frac{\partial \sigma}{\partial z} &amp;= \frac{\partial}{\partial z} \left(\frac{1}{1 + e^{-z}}\right) \\&amp;= \frac{e^{-z}}{(1 + e^{-z})^2} \\&amp;= \sigma(z)(1 - \sigma(z))\end{aligned}" role="presentation" style="text-align: center; position: relative;">∂σ∂z=∂∂z(11+e−z)=e−z(1+e−z)2=σ(z)(1−σ(z))∂σ∂z=∂∂z(11+e−z)=e−z(1+e−z)2=σ(z)(1−σ(z)) \begin{aligned}\frac{\partial \sigma}{\partial z} &= \frac{\partial}{\partial z} \left(\frac{1}{1 + e^{-z}}\right) \\&= \frac{e^{-z}}{(1 + e^{-z})^2} \\&= \sigma(z)(1 - \sigma(z))\end{aligned}<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mtable rowspacing="0.25em" columnalign="right left" columnspacing="0em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mfrac><mrow><mi mathvariant="normal">∂</mi><mi>σ</mi></mrow><mrow><mi mathvariant="normal">∂</mi><mi>z</mi></mrow></mfrac></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mfrac><mi mathvariant="normal">∂</mi><mrow><mi mathvariant="normal">∂</mi><mi>z</mi></mrow></mfrac><mrow><mo fence="true">(</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><msup><mi>e</mi><mrow><mo>−</mo><mi>z</mi></mrow></msup></mrow></mfrac><mo fence="true">)</mo></mrow></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mfrac><msup><mi>e</mi><mrow><mo>−</mo><mi>z</mi></mrow></msup><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msup><mi>e</mi><mrow><mo>−</mo><mi>z</mi></mrow></msup><msup><mo stretchy="false">)</mo><mn>2</mn></msup></mrow></mfrac></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mi>σ</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>1</mn><mo>−</mo><mi>σ</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></mstyle></mtd></mtr></mtable><annotation encoding="application/x-tex">\begin{aligned}
\frac{\partial \sigma}{\partial z} &#x26;= \frac{\partial}{\partial z} \left(\frac{1}{1 + e^{-z}}\right) \\
&#x26;= \frac{e^{-z}}{(1 + e^{-z})^2} \\
&#x26;= \sigma(z)(1 - \sigma(z))
\end{aligned}</annotation></semantics></math>∂z∂σ​​=∂z∂​(1+e−z1​)=(1+e−z)2e−z​=σ(z)(1−σ(z))​So, the partial derivative of the sigmoid function with respect to z" role="presentation" style="position: relative;">zzz is &#x03C3;(z)(1&#x2212;&#x03C3;(z))" role="presentation" style="position: relative;">σ(z)(1−σ(z))σ(z)(1−σ(z))\sigma(z)(1 - \sigma(z)).z" role="presentation" style="position: relative;">zzz is $\sigma(z
물론 시그모이드 함수는 다음과 같이 쓸 수 있습니다:

&#x3C3;(z)=11+e&#x2212;z \sigma(z) = \frac{1}{1 + e^{-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}}<math xmlns="http://www.w3. org/1998/Math/MathML" display="block"><시맨틱><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><msup><mi>e</mi><mrow><mo>-</mo><mi>z</mi></mrow></msup></mrow></mfrac></mrow><annotation encoding="application/x-. tex">\sigma(z) = \frac{1}{1 + e^{-z}}</annotation></semantics></math>σ(z)=1+e-z1where z" role="presentation" style="position: 상대;">zzz는 입력에 바이어스를 더한 가중치 합계입니다.

z에 대한 시그모이드 함수의 편미분은 다음과 같이 미분 몫 규칙을 사용하여 구할 수 있습니다." role="presentation" style="위치: 상대적;">zzz는 다음과 같이 미분 몫 규칙을 사용하여 구할 수 있습니다.

&#x2202;&#x3C3;&#x2202;z=&#x2202;&#x2202;z(11+e&#x2212;z)=e&#x2212;z(1+e&#x2212;z)2=&#x3C3;(z)(1&#x2212;&#x3C3;(z)) \begin{aligned}\frac{\부분 \시그마}{\부분 z} &amp;= \frac{\부분}{\부분 z} \왼쪽(\frac{1}{1 + e^{-z}}\right) \\&amp;= \frac{e^{-z}}{(1 + e^{-z})^2} \\&amp;= \sigma(z)(1 - \sigma(z))\end{aligned}" role="presentation" style="text-align: center; position: relative;">∂σ∂z=∂∂z(11+-e-z)=e-z(1+-e-z)2=σ(z)(1-σ(z))∂σ∂z=∂∂z(11+-e-z)=e-z(1+-e-z)2=σ(z)(1-σ(z)) \begin{aligned}\frac{\partial \sigma}{\partial z} &= \frac{\partial}{\partial z} \left(\frac{1}{1 + e^{-z}}\right) \\&= \frac{e^{-z}}{(1 + e^{-z})^2} \\&= \sigma(z)(1 - \sigma(z))\end{aligned}<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mtable rowspacing="0. 25m" columnalign="오른쪽 왼쪽" columnspacing="0em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mfrac><mrow><mi mathvariant="normal">∂</mi><mi>σ</mi></mrow><mrow><mi mathvariant="normal">∂</mi><mi>z</mi></mrow></mfrac></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mfrac><mi mathvariant="normal">∂</mi><mrow><mi mathvariant="normal">∂</mi><mi>z</mi></mrow></mfrac><mrow><mo fence="true">(</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><msup><mi>e</mi><mrow><mo>-</mo><mi>z</mi></mrow></msup></mrow></mfrac><mo fence="true">)</mo></mrow></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mfrac><msup><mi>e</mi><mrow><mo>-</mo><mi>z</mi></mrow></msup><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msup><mi>e</mi><mrow><mo>-</mo><mi>z</mi></mrow></msup><msup><mo stretchy="false">)</mo><mn>2</mn></msup></mrow></mfrac></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mi>σ</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><mi>σ</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></mstyle></mtd></mtr></mtable><annotation encoding="application/x-. tex">\begin{aligned}
\frac{\partial \sigma}{\partial z} &#x26;= \frac{\partial}{\partial z} \왼쪽(\frac{1}{1 + e^{-z}}\right) \\
&#x26;= \frac{e^{-z}}{(1 + e^{-z})^2} \\
&#x26;= \sigma(z)(1 - \sigma(z))
\end{aligned}</annotation></semantics></math>∂z∂σ=∂z∂(1+e-z1)=(1+e-z)2e-z=σ(z)(1-σ(z))따라서 z에 대한 시그모이드 함수의 편미분은" role="presentation" style="position": 상대적;">zzz는 &#x03C3;(z)(1&#x2212;&#x03C3;(z))" role="presentation" style="위치: 상대적;">σ(z)(1-σ(z))σ(z)(1-σ(z))\sigma(z)(1 - \sigma(z)) 입니다. z" role="presentation" style="위치: 상대적;">zzz는 $\sigma(z)입니다.