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Page 365, missing negative sign in the calculation of partial derivative of activation unit with respect to z(out) #140

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acmoudleysa opened this issue Aug 16, 2023 · 0 comments

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@acmoudleysa
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$$ \frac{\partial a_1^{(out)}}{\partial z_1^{(out)}} = \frac{\partial}{\partial z_1^{out}}\cdot \frac{1}{1+e^{z_1^{(out)}}} = ... = \left(\frac{1}{1+e^{z_1^{(out)}}}\right) \cdot \left(1-\frac{1}{1+e^{z_1^{(out)}}}\right)$$

In this equation, the exponents should have negative sign

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