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The question is about stationary point of a least squares being identical to the stationary point of an error function as stated in: Computer Vision: Algorithms and Applications by Szelinski et al. p.318

The premise is:

For a cost function with robust error function $h(e_i)$ $$E(\delta p)=\sum\limits_{i=0}^nh(\|r\|)$$

Where h is a robust penalty function and r is the residual. $$w(x) = \frac{h'(x)}{x}$$

$$E_w(p)=\sum\limits_{i=0}^nw(\|r\|)\|r\|^2,$$

$$ \frac{\delta h(\|r\|)}{\delta p} = w(\|r\|)\|r\| \equiv \frac{\delta E_w}{\delta p}$$

This is what is stated in the text. However, when I try to derive this I end up with:

$$w(\|r\|)\|r\| + h''(\|r\|)\|r\|$$ due to the product rule.

So in this case, is the second derivative just ignored?

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