Differentiate the norm of a function

Question

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.

Answer 1

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.

Answer 2

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