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I'm quite unsure about how to deal with differentiation of absolute functions, and their continuity.

For example, the question I was dealing with was the following: $$ f(x) = \frac{x}{1 + |x|}$$

According to me, the equation is continuously differentiable everywhere, given there's no sharp point in its graph. (The graph which I constructed below)

enter image description here

Apparently, though, the derivative has a point of dicontinuity at 0. (According to the solutions in my problem set). I can't intuitively understand why that would be.

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  • $\begingroup$ your function $f = x + \cdots$ for $x$ close to $0.$ $f$ behaves like $x$ when $x$ is close to zero. $\endgroup$
    – abel
    Apr 9 '15 at 14:29
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There is no discontinuity at zero, and the derivative is continuous there as well. But to compute the derivative at zero you need to split the function up into the two domains $x\le 0$ and $x\ge 0$: \begin{equation*} f(x) = \begin{cases} \frac{x}{1+x},&x\ge 0 \\ \frac{x}{1-x},&x<0.\end{cases} \end{equation*} It is easy to differentiate $f$ away from zero, but at zero you must compute the limits from the left and right and show that they are equal; if you do so, you will see that they are both $1$, so that $f$ is differentiable at $0$ and $f'(0) = 1$.

What is true, however (as you can see if you plot $f'(x)$) is that $f'(x)$ is not differentiable at $0$.

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  • $\begingroup$ Thanks! This makes sense to me. $\endgroup$
    – dexter
    Apr 9 '15 at 14:14
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Your function is given by $$ f(x) = \begin{cases} \frac x{1+x} & x \ge 0 \\ \frac x{1-x} & x \le 0 \end{cases}. $$ $f$ is obviously differentiable on $\mathbf R\setminus \{0\}$, with $$ f'(x) = \begin{cases} \frac 1{(1+x)^2} & x >0 \\ \frac 1{(1-x)^2} & x < 0 \end{cases} $$ moreover $f_+'(0) = 1 = f_-'(0)$, hence $f$ is continuously differentiable with $$ f'(x) = \begin{cases} \frac 1{(1+x)^2} & x \ge 0 \\ \frac 1{(1-x)^2} & x \le 0 \end{cases}. $$

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