How to calculate $ \left| \sin x \right| $ derivative in a more elegant way?

Question

I am trying to calculate the derivative of $\left| \sin x \right| $

Given the graphs, we notice that the derivative of $\left| \sin x \right|$ does not exist for $x= k\pi$.

Graph for $\left|\sin x\right|$:

We can rewrite the function as

$\left| \sin(x) \right| = \left\{ \begin{array}{ll} \sin(x),& 2k\pi < x < (2k+1)\pi \\ -\sin(x), & \text{elsewhere} \\ \end{array} \right. $

Therefore calculate its derivative as:

$(\left| \sin(x) \right|)^{'} = \left\{ \begin{array}{ll} \cos(x),& 2k\pi < x < (2k+1)\pi \\ -\cos(x), & \text{elsewhere} \\ \end{array} \right. $

Is there a way to rewrite this derivative, in a more elegant way (as a non-branch function) $(\left| \sin(x) \right|)^{'} = g(x)$?

Answer 1

The better way, for me, is as follows: $$f(x)=\left|\sin(x)\right|=\sqrt{\sin^2(x)}$$ Now, differentiate both sides to get $$f'(x)=\frac{1}{2\sqrt{\sin^2(x)}}\cdot2\sin(x)\cos(x)=\frac{\sin(2x)}{2\left|\sin(x)\right|}$$ Therefore, $$\left(\left|\sin(x)\right|\right)'=\frac{\sin(2x)}{2\left|\sin(x)\right|}, \ \ \ \ x \neq k\pi, k\in \mathbb{Z}$$

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Appendum: This approach can easily be extended to a general case of finding $\left(|f(x)|\right)'$.

First, we rewrite $$|f(x)| = \sqrt{f^2(x)}$$ Then, repeating the work above: $$|f(x)|' = \frac{1}{2\sqrt{f^2(x)}}[2f(x)f'(x)] = \frac{f(x)}{|f(x)|}f'(x)$$ we get $$\boxed{|f(x)|' = \frac{f(x)}{|f(x)|}f'(x)}$$

Answer 2

Apparently other answers missed the following direct approach: we have $$|x|' = \frac{x}{|x|}$$ for $x\ne 0$.

So, for example, VIVID's formula $|f(x)|' = \frac{f(x)}{|f(x)|} f'(x)$ is just the chain rule.

Answer 3

I think an even nicer way of thinking about it is in terms of the sign of $\sin(x)$ applied to the derivative of $\sin(x)$. A function $f$ that returns the sign of another function $g$ is: $f(x)=\frac{\lvert(g(x))\rvert}{g(x)}$ .
So quite simply $\lvert \sin(x) \rvert'=\frac{\lvert \sin(x)) \rvert}{\sin(x)}\cdot\cos(x)$

This is essentially just $\cos(x)$ but reflected across the axis at the intervals where $\sin(x)$ is negative, that is: $\cos(x)$ for $x \in [0,\pi]$ and $-\cos(x)$ for $x \in [\pi,2\pi]$ and so on. The reason for this is not difficult to see: when we take the absolute value of a function, then the gradients of the points belonging to the intervals that were flipped, here where $x \in [\pi,2\pi]$, are also flipped. So what ends up happening is that the derivative essentially stays the same, but is flipped on the intervals where the original function was flipped.