Is this function differentiable on $\mathbb{R}$?

Question

Here is my prood to see if this function is differentiable over $\mathbb{R}$.

$$f(x) = \vert x\vert x$$

\begin{equation*} \begin{split} \lim_{h\to 0} \dfrac{(x+h)\vert x+h \vert - x\vert x \vert}{h} & = \lim_{h\to 0} \dfrac{x\vert x+h \vert + h \vert x + h\vert - x\vert x \vert }{h} \\\\ & = \lim_{h\to 0} \dfrac{x\vert x \vert + x\vert h\vert + h \vert x \vert + h\vert h \vert - x\vert x \vert}{h} \\\\ & = \lim_{h\to 0} x \dfrac{\vert h \vert}{h} + \vert x \vert + \vert h \vert \end{split} \end{equation*}

Now the last term is zero. The middle term is just $\vert x \vert $ and the first one now depends on the direction:

$$\lim_{h\to 0^+} x \dfrac{\vert h \vert}{h} = x$$

$$\lim_{h\to 0^-} x \dfrac{\vert h \vert}{h} = -x$$

Consequently, the limits are different (here $x$ can be whatever number), hence the function exhibits an angular point at every $x$ in the domain.

This concludes the function is not differentiable in $\mathbb{R}$.

Now it looks weird to me, maybe I'm right or perhaps I'm wrong.

Asking you for help in case!

Answer 1

Following the comments you received, we consider the three cases:

• $x >>0$

• $x << 0$

• $x = 0$

Case 1, $x >>0$

Here then $\vert x+ h \vert = x + h$ hence

\begin{equation} \begin{split} \lim_{h\to 0} \dfrac{(x+h)^2 - x^2}{h} & = \lim_{h\to 0} \dfrac{2hx + h^2}{h} \\\\ & = \lim_{h\to 0} 2x + h \longrightarrow 2x \end{split} \end{equation}

Case 2, $x << 0$

Here then $\vert x+ h \vert = -(x + h)$ hence

\begin{equation} \begin{split} \lim_{h\to 0} \dfrac{-(x+h)^2 + x^2}{h} & = \lim_{h\to 0} \dfrac{-2hx - h^2}{h} \\\\ & = \lim_{h\to 0} -2x - h \longrightarrow -2x \end{split} \end{equation}

Case 3, $x = 0$

Here then $\vert x+ h \vert = h$ hence

\begin{equation} \begin{split} \lim_{h\to 0} \dfrac{h\vert h \vert}{h} & = \lim_{h\to 0} \vert h\vert \\\\ & \longrightarrow 0 \end{split} \end{equation}

Whence the function $f(x) = x\vert x\vert$ is differentiable over all $\mathbb{R}$.

Answer 2

Firstly, use that $x\mapsto|x|$ and $x\mapsto x$ are differentiable for every $x\neq0$ to conclude that $f$ is differentiable for $x\neq0$.

To see derivability at $0$ consider $$\frac{f(h)-f(0)}{h}$$ and see that it tends to zero as $h\to 0$.