Derivative of $$$- \epsilon_{k} + z$$$ with respect to $$$\epsilon_{k}$$$
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Find $$$\frac{d}{d\epsilon_{k}} \left(- \epsilon_{k} + z\right)$$$.
Solution
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{d\epsilon_{k}} \left(- \epsilon_{k} + z\right)\right)} = {\color{red}\left(- \frac{d}{d\epsilon_{k}} \left(\epsilon_{k}\right) + \frac{dz}{d\epsilon_{k}}\right)}$$Apply the power rule $$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}^{n}\right) = n \epsilon_{k}^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}\right) = 1$$$:
$$- {\color{red}\left(\frac{d}{d\epsilon_{k}} \left(\epsilon_{k}\right)\right)} + \frac{dz}{d\epsilon_{k}} = - {\color{red}\left(1\right)} + \frac{dz}{d\epsilon_{k}}$$The derivative of a constant is $$$0$$$:
$${\color{red}\left(\frac{dz}{d\epsilon_{k}}\right)} - 1 = {\color{red}\left(0\right)} - 1$$Thus, $$$\frac{d}{d\epsilon_{k}} \left(- \epsilon_{k} + z\right) = -1$$$.
Answer
$$$\frac{d}{d\epsilon_{k}} \left(- \epsilon_{k} + z\right) = -1$$$A