# Why doesn't my derivation of the arcsec derivative formula not work?

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.

• 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.) Jan 25, 2021 at 20:09
• 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. Jan 25, 2021 at 20:12
• In that case, why do you think it should be positive? I'm just pointing out one way to resolve your contradiction. Jan 25, 2021 at 20:13

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 $$x>0$$.)

• [+1] Nice answer,
– user822140
Jan 25, 2021 at 20:15
• got it thank you Jan 25, 2021 at 20:28
• [+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? Jan 25, 2021 at 20:44
• @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. Jan 25, 2021 at 20:58
• Oh, I see! Thanks for the answer, have a good day! 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}