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I would like to take the first derivative of a norm like this. I got the answer, but I am here to double check. Thank you.

$$\frac{d}{dx'}\left\{\exp\left[-\frac{\|x'-x\|^2} \theta \right]\right\},$$ where $\theta$ is a constant. $x'$ and $x$ are $1 \times p$ row vector, respectively.

Michael Hardy
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Sophia
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  • You'll get more useful answers if you show what your answer is and how you got it. 2) Hint: $| x' - x |^2 = (x' - x)^T (x' -x)$
    • – erfink May 12 '17 at 02:30