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How should I differentiate the norm of a function? I mean, how can I get the first and second derivatives of something like: $$||\alpha(s)||^2$$ I know that I have to use the chain rule, but I am struggling with it. Thanks.

dmtri
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southernKid33
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2 Answers2

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Without details of the norm, define $n$ via $n(x) = ||x||$. Then \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} ||f(x)||^2 &= \frac{\mathrm{d}}{\mathrm{d}x} n(f(x))^2 \\ &= 2n(f(x)) \cdot n'(f(x)) \cdot f'(x) \\ &= 2||f(x)|| n'(f(x)) f'(x) \text{.} \end{align*} If you have a particular norm in mind, you should be able to use its derivative for the middle factor.

Eric Towers
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  • The euclidean norm. Actually, I am trying to figure out why this guy got this derivative in this answer: https://math.stackexchange.com/a/496013/566632 – southernKid33 Sep 13 '19 at 03:00
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    You could have saved us both time if you had asked that directly. $\langle u,v \rangle'' = (\langle u',v \rangle + \langle u, v' \rangle)' = \langle u'',v \rangle + 2\langle u',v' \rangle + \langle u, v'' \rangle$. Set $u = v = \alpha(s)$ and notice the middle term is $2\langle \alpha'(s), \alpha'(s) \rangle = 2||\alpha'(s)||^2$. – Eric Towers Sep 13 '19 at 03:06
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    Thank you so much. I am so sorry, I thought I was not able to ask specific questions like that. I really appreciate it – southernKid33 Sep 13 '19 at 03:07
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Is this a vector valued function?

If this is just the absolute value of a regular function, then $\displaystyle |x|=\sqrt{x^2}$ and so $\displaystyle \frac{d}{dx}|f(x)|=\frac{d}{dx}\sqrt{f(x)^2}=\frac{2f(x)f'(x)}{2\sqrt{f(x)^2}}=\frac{f'(x)f(x)}{|f(x)|}=f'(x)\cdot\text{sign}(f(x))$.

On the other side, if it is a vector valued function $f(x)=<f_1(x),f_2(x),f_3(x)>$, then you have $\displaystyle |f(x)|=\sqrt{f_1(x)^2+f_2(x)^2+f_3(x)^2}$ and follow a similar approach.

Here, you would get $\displaystyle \frac{d}{dx}|f(x)|=\frac{2f_1(x)f_1'(x)+2f_2(x)f_2'(x)+2f_3(x)f_3'(x)}{2\sqrt{f_1(x)^2+f_2(x)^2+f_3(x)^2}}=\frac{f_1(x)f'_1(x)+f_2(x)f_2'(x)+f_3(x)f'_3(x)}{|f(x)|}$

Saketh Malyala
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