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How to justify, the following function is differentiable at origin or not?

$f(x) = \tan^{-1}\frac{1}{|x|}$ if $x \ne 0$, $f(x) = \frac{\pi}{2}$ if $x = 0$.

Even though mod x is not behaves well at the origin, since we are composing the mod x with the nice function $\tan^{-1}(x)$, I am guessing that above defined $f$ is differentiable at 0. But how to show that rigorously?

I think my guess is correct.

Thanks in Advance.

mathlove
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GA316
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2 Answers2

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First note that $$ f(x)=\tan^{-1}\left(\frac1{|x|}\right)=\frac\pi2-\tan^{-1}(|x|) $$ Then $$ \lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{-\tan^{-1}(h)}{h}=-1 $$ and $$ \lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{-\tan^{-1}(-h)}{h}=1 $$ Since the two limits don't match, the function is not differentiable at $x=0$.

Here is a plot that shows the slope of $1$ to the left and the slope of $-1$ to the right:

$\hspace{3.5cm}$enter image description here

robjohn
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \color{#66f}{\large\totald{\arctan\pars{1/\verts{x}}}{x}} ={1 \over \pars{1/\verts{x}}^{2} + 1}\, \pars{-\,{1 \over \verts{x}^{2}}}\sgn\pars{x} =\color{#66f}{\large-\,{\sgn\pars{x} \over x^{2} + 1}}\,,\qquad x \not=0 $$

Note that $$ \lim_{x\ \to\ 0^{-}}\totald{\arctan\pars{1/\verts{x}}}{x} = 1 \quad\mbox{and}\quad \lim_{x\ \to\ 0^{+}}\totald{\arctan\pars{1/\verts{x}}}{x} = -1 $$

Namely, $$\left.\totald{\arctan\pars{1/\verts{x}}}{x}\right\vert_{\,x\ =\ 0^{-}} \not=\left.\totald{\arctan\pars{1/\verts{x}}}{x}\right\vert_{\,x\ =\ 0^{+}} $$

such that $\ds{\left.\totald{\arctan\pars{1/\verts{x}}}{x}\right\vert_{\,x\ =\ 0}}$ $\ds{\underline{\tt\mbox{doesn't exist}}}$.

Felix Marin
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  • Since this tells what the derivative is away from the point in question, it would be nice to mention that if the limit of the derivative from the left of $x_0$ and the limit of the derivative from the right of $x_0$ exist and are not equal, then the derivative does not exist at $x_0$ (say using the Mean Value Theorem or something similar). – robjohn Sep 22 '14 at 12:04
  • @robjohn It's pretty clear from the result. – Felix Marin Sep 23 '14 at 16:49
  • @robjohn I add the "Note that...". Thanks. – Felix Marin Sep 23 '14 at 16:57
  • I meant that because $$\frac{f(x)-f(0)}{x-0}=f'(\xi)\text{ for some $\xi$ between $x$ and $0$}$$ we get that the left derivative at $0$ is $1$ and the right derivative at $0$ is $-1$. Perhaps that is what you meant by $$\left.\frac{\mathrm{d}\arctan(1/|x|)} {\mathrm{d}x},\right|{x=0^-} \quad\text{and}\quad \left.\frac{\mathrm{d}\arctan(1/|x|)} {\mathrm{d}x},\right|{x=0^+}$$ However, those seem to indicate values just to the left and right of $0$, which is different. – robjohn Sep 23 '14 at 17:20