If only continuous functions can be differentiable, then how can the tangent function $\tan$ be differentiated, even though $\tan$ is not a continuous function?

  • $\begingroup$ $\tan x$ is not differentiable on $\Bbb R$. $\endgroup$ – Ruben Jul 2 at 9:39
  • $\begingroup$ $\tan$ is continuous. All elementary functions are continuous everywhere. $\endgroup$ – Alexey Jul 2 at 9:48
  • $\begingroup$ @Ruben, $\tan$ is not defined on $\mathbb{R}$. "$\tan$ is not differentiable on $\mathbb{R}$" is a bit like saying that the square root function is not differentiable on $\mathbb{Z}[X, Y]$. $\endgroup$ – Alexey Jul 2 at 9:53
  • $\begingroup$ @alexey It's more like the square root function is not differentiable on $\Bbb R$, which is true. $\endgroup$ – Ruben Jul 2 at 9:58
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    $\begingroup$ @Alexey Ok sure, of course the statement is kind of silly. I was just trying to stimulate the person asking the question into thinking carefully about the domain of the function and its points of discontinuity. $\endgroup$ – Ruben Jul 2 at 10:06

The tangent function is continuous. I suppose that you are confused by the fact that it has singularities at $k\pi+\frac{\pi}{2}$, $k\in\Bbb Z$, but these points are simply not in the domain of the function.

To be more precise: The function

$$ \tan\colon \Bbb R \setminus \{k\pi+\frac{\pi}{2}: k\in\Bbb Z\} \to \Bbb R$$

is continuous and indeed even differentiable.


Actually, $\tan$ is a continuous (and differentiable) function. Keep in mind that its domain is $\mathbb R\setminus\left\{\frac\pi2+k\pi\,\middle|\,k\in\mathbb Z\right\}$. Since it is the quotient of two differentiable functions, it is differentiable.


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