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$\tan x$ is not differentiable at $(2n + 1)90$ points, which means function itself is not differentiable. So, why does its differentiation $\sec^2(x)$ exists?

Andrew D. Hwang
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user2409011
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    Now, why would you downvote it? My question could be dumb but anybody can have doubts. – user2409011 Feb 14 '16 at 11:49
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    The expression $\tan x$ is not a function, $\tan$ is. And $\tan$ is differentiable at every point in its domain, so it is differentiable. – Git Gud Feb 14 '16 at 11:50
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    $\sec^2x$ is not defined at these points either. – Bernard Feb 14 '16 at 11:50
  • @GitGud tan is not differentiable at 90 degree, since it is not continuous. Am i wrong? – user2409011 Feb 14 '16 at 11:51
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    @user2409011 tan is not defined at 90 degrees. So, there is no question of it being continuos or differentiable. It's just not defined. – Win Vineeth Feb 14 '16 at 11:52
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    @user2409011 Differentiability only makes sense in points of the domain of the given function. It doesn't make sense to talk about differentiability or non-differentiability of $\tan$ at $90º$. It's not that $\tan$ isn't differentiable there, it's that that concept doesn't even make sense. – Git Gud Feb 14 '16 at 11:54
  • @user2409011 Perhaps this answer and the comments below it will help you understand. – Git Gud Feb 14 '16 at 11:57
  • This is a duplicate question and has been asked many times. – Max Payne Feb 14 '16 at 12:09

2 Answers2

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You derive $\tan(x)$ on its domain, not on any point of $\mathbb{R}$!

Also, note that $\sec^2(x) = \frac{1}{\cos^2 x}$ is not defined at $x= \frac{\pi}{2} + k\pi$, which is coherent with the fact that $\tan(x)$ isn't too.

AnalysisStudent0414
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Differentiation here is defined only on those points of $\tan x$ which are differentiable. In the regions where the function is differentiable, It has a derivative equal to $\sec^2(x)$

Win Vineeth
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