$$$\epsilon_{k} + z$$$ 對 $$$\epsilon_{k}$$$ 的導數
您的輸入
求$$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k} + z\right)$$$。
解答
和/差的導數等於導數的和/差：
$${\color{red}\left(\frac{d}{d\epsilon_{k}} \left(\epsilon_{k} + z\right)\right)} = {\color{red}\left(\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}\right) + \frac{dz}{d\epsilon_{k}}\right)}$$套用冪次法則 $$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}^{n}\right) = n \epsilon_{k}^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}\right) = 1$$$：
$${\color{red}\left(\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}\right)\right)} + \frac{dz}{d\epsilon_{k}} = {\color{red}\left(1\right)} + \frac{dz}{d\epsilon_{k}}$$常數的導數為$$$0$$$：
$${\color{red}\left(\frac{dz}{d\epsilon_{k}}\right)} + 1 = {\color{red}\left(0\right)} + 1$$因此，$$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k} + z\right) = 1$$$。
答案
$$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k} + z\right) = 1$$$A