1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $\tan x$ 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
  • 2
    $\begingroup$ @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
6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.