How to find the derivative of $|f(x)|$

Question

The original question was to find domain of derivative of $y=|\arcsin(2x^2−1)|$.

First method
My attempt was to break $y$ into intervals ,i.e., where $\arcsin(2x^2−1)\geq 0$ and where $\arcsin(2x^2−1)<0$,and then differentiate the resulting function and find its domain. But my teacher said that this is wrong. He argued that this is like putting values to a function(which makes it a constant) and then differentiating.For eg- If we put $x=0$ and then differentiate we will get $y'=0$ which may be wrong.Is my teacher correct?

Second method
I have learnt that by putting $u=\arcsin(2x^2−1)$ we get $\frac{dy}{dx}=\frac{d|u|}{du} \cdot \frac{du}{dx}$ and $\frac{d|u|}{du}=\frac{|u|}{u}$ (but I failed to understand this step) and using this we can find the required derivative.

If I use second method then the domain does not include $x=0$ , and if i use first it includes $x=0$. So is my method (first) totally wrong?

Answer 1

First off, $y < 0$ is nonsense because $y$ is an absolute value of something, thus nonnegative.

The proper way to handle such things with cases is

• You must find the "critical points" of the function. Those are the ones where the absolute value is $0$, i.e. $$\arcsin(2x^2-1) = 0$$ Here these are $x = \pm \frac1{\sqrt 2}$.
• You then compute the derivatives on the open intervals where you can explicitly decide the sign of your expression. The natural domain of your function is $[-1,1]$ (why?), so the intervals are $[-1, -\frac1{\sqrt 2}), (-\frac1{\sqrt 2}, \frac1{\sqrt 2}), (\frac1{\sqrt 2}, 1]$ (note how I am excluding the critical points)
• On these intervals, you can compute the derivative as usual
• Finally, check if you can "patch together" the three parts continuously. If so, the derivative at those critical points exists and is equal to the left and right limit. If not, it doesn't exist. An example where this doesn't work is $|x|$ (Derivative $-1$ for $x<0$ and $+1$ for $x>0$), but it does work with $|x^3|$.