Which are differentiable for all x?

Question

$F_1:$

\begin{cases} x^2/|x|, & \text{$x \neq 0$} \\ 0, & \text{$x=0$} \end{cases}

$F_2$:$$|\sin(x)|^2$$

$F_3$:$$|\cos(x)|$$

I know that the absolute value of $\cos (x)$ would not be differentiable, but the absolute value of $\sin (x)^2$ would be. Would $F_1$ be differentiable, since the corner is filled in with $(0,0)$?

Answer 1

$F_2$ is differentiable for all $x$.

Reason: In order for the function to be differentiable, the derivative must exist and the function must be continuous.

Since $x^2$ is positive for all $x$, we can rewrite $|\sin (x)|^2$ as just plain $\sin(x)^2$. Then the derivative is $$\dfrac {dy}{dx} = 2 \sin(x) \cos(x) = \sin (2x)$$

Then, we determine if the function is continuous (i.e. there are no values that would render the function invalid). Since $2 \sin (x) \cos (x) = \sin (2x)$ is valid for all values of $x$, it is continuous, so the original function $|\sin(x)|^2$ is also continuous for all $x$. Hence $F_2$ is differentiable for all $x$.

Answer 2

There's a typo in your $F_1$, it's $x\neq 0$, otherwise it wouldn't be a function. $F_1$ is not differentiable at $x=0$

$F_1(x)=\begin{cases} x^2/|x| & x \neq 0 \\ 0 & x=0 \end{cases}$

but $ x^2 =|x|^2$ then $\frac{|x|^2}{|x|}=|x|$

and if $x=0$ then $|x|=0$

Then $F_1$ actually is a strange wording for $F_1(x)= |x|$

And you know $|x|$ isn't differentiable at $x=0$ because

$\frac{F_1(x+\Delta x)-F_1(x)}{\Delta x}\underset{x\rightarrow 0^+}{\longrightarrow}-1$ but $\frac{F_1(x+\Delta x)-F_1(x)}{\Delta x}\underset{x\rightarrow 0^-}{\longrightarrow}1$ So there's no limit, and that limit is the definition of the derivative and one-variable functions are differentiable at a point if the derivative exists at the same point.

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For $F_3$ your function isn't differentiable at $\frac{\pi}{2}+k\pi\quad k\in\mathbb{Z}$

To prove that, take two different cases $\frac{\pi}{2}+2k\pi$ and $\frac{3\pi}{2}+2k\pi$

and notice your function can be written as:

$F_3=|\cos(x)|= \begin{cases} -\cos(x) & \cos(x) \leq 0 \\ \cos(x) & \cos(x) > 0 \end{cases} = \begin{cases} -\cos(x) & x \in \underset{k\in \mathbb{Z}}{\cup} [\frac{\pi}{2}+2k\pi, \frac{3\pi}{2}+2k\pi] \\ \cos(x) & x \in \underset{k\in \mathbb{Z}}{\cup} (\frac{3\pi}{2}+2k\pi, \frac{5\pi}{2}+2k\pi) \end{cases} $

Then, at the interior of each interval the derivative is either $-\sin(x)$ or $\sin(x)$

$F'_3(x) = \begin{cases} \sin(x) & x \in \underset{k\in \mathbb{Z}}{\cup} (\frac{\pi}{2}+2k\pi, \frac{3\pi}{2}+2k\pi) \\ -\sin(x) & x \in \underset{k\in \mathbb{Z}}{\cup} (\frac{3\pi}{2}+2k\pi, \frac{5\pi}{2}+2k\pi) \end{cases}$

And that isn't continuous at $\frac{\pi}{2}+k\pi$

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$F_2$ is differentiable at all x as others have explained.