We would like to find the derivative of $\operatorname{arcsec}(x) = y$. Rearranging this, we get $x = \sec(y)$. Taking the derivative of both sides, we get $1 = \sec(y)\tan(y)y^\prime$. Thus, $$y^\prime = \frac{1}{\sec(y)\tan(y)}.$$ Now, draw a right triangle with an acute angle $y$, hypotenuse $x$, and a side adjacent to angle $y$ with length $1$. By the Pythagorean theorem, the side opposite angle $y$ is equal to $\sqrt{x^2-1}$. Thus, we have $x = \sec(y)$, which we already knew and got from the triangle, and $$\tan(y) = \frac{1}{\sqrt{x^2-1}}$$ from the triangle we drew. We substitute these values in to our derivative expression to get $$y' = \frac{1}{x\sqrt{x^2-1}}.$$

However, $y^\prime$ should equal $$\frac1{|x|\sqrt{x^2-1}}$$ (note the absolute value). Since the difference occurred in the $|x|$ part, that means I must've done something wrong when I said $\sec(y) = x$, but I do not know why this assumption is wrong. Please correct my proof.

  • 1
    $\begingroup$ If your hypotenuse $x$ has a negative length, you have to give the other side a negative length to compensate, making it $-\sqrt{x^2-1}$. (I can't make this rigorous, but I think it's the source of your problem.) $\endgroup$
    – TonyK
    Jan 25, 2021 at 20:09
  • $\begingroup$ Why do I have to make the other side a negative length? By pythagoras we have a^2 + b^2 = c^2, so a,b,c can be negative, or positive, while satisfying this equation. $\endgroup$
    – Some Guy
    Jan 25, 2021 at 20:12
  • $\begingroup$ In that case, why do you think it should be positive? I'm just pointing out one way to resolve your contradiction. $\endgroup$
    – TonyK
    Jan 25, 2021 at 20:13

2 Answers 2


The sloppy reasoning occurs with the right triangle you wrote. You must be careful about the range (values) of the arcsec function. $y=\text{arcsec}(x)$ can lie either in $[0,\pi/2)$ or in $(\pi/2,\pi]$. (These correspond, respectively, to $x>0$ and $x<0$.)

$\tan(y)<0$ when $y\in (\pi/2,\pi]$, and so $\tan(y)=-\sqrt{x^2-1}$ in that event. Working with your formula for $y'$, we note that when $x<0$, $$\sec(y)\tan(y)= x(-\sqrt{x^2-1})= (-x)\sqrt{x^2-1} =|x|\sqrt{x^2-1},$$ and this explains the formula. To be honest, it's a bit sneaky to move the negative to the other term, but it allows us to write a single formula, rather than writing down cases. (There's no problem, of course, when $x>0$.)

  • $\begingroup$ [+1] Nice answer, $\endgroup$
    – user822140
    Jan 25, 2021 at 20:15
  • $\begingroup$ got it thank you $\endgroup$
    – Some Guy
    Jan 25, 2021 at 20:28
  • $\begingroup$ [+1], nice. Can you explain what you mean with "it's a bit sneaky to move the negative to the other term"? Why is it sneaky? $\endgroup$
    – ZaWarudo
    Jan 25, 2021 at 20:44
  • $\begingroup$ @ZaWarudo I was not meaning anything too serious, certainly nothing pejorative. Obviously, there's nothing too creative about reassociating the $-1$ factor. But a beginning calculus student might not think to do this. $\endgroup$ Jan 25, 2021 at 20:58
  • $\begingroup$ Oh, I see! Thanks for the answer, have a good day! $\endgroup$
    – ZaWarudo
    Jan 25, 2021 at 21:10

Ted Shifrin has already explained the problem with your derivation, but if you're looking for a 'safer' way of finding the derivative then consider this: if $\DeclareMathOperator{\arcsec}{arcsec} y = \arcsec x$, then using the identity $\arcsec x = \arccos 1/x$, we see that \begin{align} \frac{dy}{dx}=\frac{d}{dx}\left(\arccos 1/x\right) &= -\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^2}} \cdot -\frac{1}{x^2} \\[4pt] &= \frac{1}{x^2\sqrt{1-\frac{1}{x^2}}} \, . \end{align}


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