I want to determine all the points where $g(x) = |\sin(2x)|$ is differentiable.
A function is differentiable at a point if the left and right limits exist and are equal.
So it follows that $g(x)$ is differentiable for all $x$ except where $g(x) = 0$. For example, the derivative of $|\sin(2x)|$ does not exist at $x=0$.
Is this correct?
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$\begingroup$It is important to make the distinction between the limit of the function and the limit of the difference quotient.
Suppose you were simply talking about the limit of the function. We have $$\lim_{x\to0}g(x)=\lim_{x\to0}|\sin2x|=0.$$
However, if you were talking about the difference quotient, the limit would not exist: $$\lim_{h\to 0^-}\frac{|\sin2(x+h)|-|\sin2x|}{h}=-2,\quad \lim_{h\to 0^+}\frac{|\sin2(x+h)|-|\sin2x|}{h}=2.$$
$\endgroup$ $\begingroup$Asking whether a function $f(x)$ is differentiable at $x=a$ is akin to asking whether the derivative $f'(x)$ is continuous at $x=a$, assuming $f(x)$ has a derivative on some interval on which $f$ is defined.
Recall that there are 3 different kinds of discontinuities at a point $x=a$: either the limit doesn't exist at all, or the limits from the left and right of $a$ exist but are not equal, or the function at hand simple isn't defined at $x=a$.
We'll differentiate $|sin(2x)|$ by applying the chain rule twice, once for the composition of $|x|$ with $sin(x)$ and then to $sin(x)$ composed with $2x$.
We obtain: $$\frac{d}{dx} |sin(2x)|=\frac{2sin(2x)cos(2x)}{|sin(2x)|}= \frac{sin(4x)}{|sin(2x)|}$$
Now, the following is not a completely rigorous argument: since $sin(4x)$ takes positive values for $x \in (0,\frac{\pi}{4}]$, and negative values for $x \in [-\frac{\pi}{4},0)$, we can deduce that the sign of the numerator in $\frac{sin(4x)}{|sin(2x)|}$ will change signs as we approach $x=0$ either from the right or from the left, and the denominator will not, since we're taking its absolute value.
Hence, the derivative of $|sin(2x)|$ is discontinuous at $x=0$, so this function isn't differentiable at $x=0$. Hope this helps.
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