5
$\begingroup$

I have been trying to derive the derivative of the arcsecant function, but I can't quite get the right answer (the correct answer is the absolute value of what I get). I first get $\frac{d}{dy}\sec(y)=\frac{\cos^2(y)}{\sin(y)}=\frac{\cos^2(\sec^{-1}(x))}{\sin(\sec^{-1}(x))}$. Then, by examining the appropriate right triangle with angle $\theta$, hypotenuse $x$, adjacent side length $1$ and opposite side length $\sqrt{x^2-1}$, we see that $\sin(\sec^{-1}(x))=\sin(\theta)=\frac{\sqrt{x^2-1}}{x}$ and $\cos(\sec^{-1}(x))=\cos(\theta)=\frac{1}{x}$. Substituting these identities into the formula for the derivative, we get $$\frac{1}{x\sqrt{x^2-1}}$$. However, the actual answer is the absolute value of that. I can't figure out where I am assuming $x$ is positive. It has been a while since I have mucked about with this kind of thing, so my apologies if this is a silly question. Wikipedia wasn't helpful on the matter.

$\endgroup$
  • 1
    $\begingroup$ In using the sohcahtoa formula, the hypothesis always has to be positive. If $x$ is negative, its length will therefore be $|x|$, not just $x$. $\endgroup$ – Christopher Carl Heckman Sep 24 '15 at 6:21
  • $\begingroup$ Oh, thanks! knew I was being thick. $\endgroup$ – irh Sep 24 '15 at 6:41
  • $\begingroup$ @irh Carl is right here, although I posted a much deeper explanation. I should note that your final answer is incorrect: the answer should be $\frac{1}{x\sqrt{x^2-1}}$ , not $\frac{1}{x\sqrt{1-x^2}}$ $\endgroup$ – Brevan Ellefsen Sep 24 '15 at 6:45
  • $\begingroup$ right, that was just a typo when writing it up. I fixed that in an edit just now. Thanks so much for your in-depth answer! $\endgroup$ – irh Sep 24 '15 at 6:47
  • $\begingroup$ @irh Sure thing, glad I could help! I've always wondered this myself... I've just accepted it as being true. In truth, I think I got as much out of the research for this question as you got from my answer! $\endgroup$ – Brevan Ellefsen Sep 24 '15 at 6:51
1
$\begingroup$

Carl is correct in his comment, it is based on the fact that a hypotenuse cannot be negative. However, we also see that if $y = \sec^{-1}(x)$, $\sec (y) = x$. We then apply the inverse function rule of derivatives to get $$\frac{d}{dx}\sec^{-1}(x) = \frac{1}{\frac{d}{dx}\sec(y)} = \frac{1}{\sec(y)\tan(y)}$$ We then reference the rule explained on this page, that the product $\sec(x)\tan(x)$ is never negative. To see this, I will quote the page:

"For, if y = arcsec x, then the angle y falls either in the first or second quadrants. When angle y falls in the first quadrant, then both sec y and tan y are positive. Therefore their product is positive. When angle y falls in the second quadrant, sec y and tan y are both negative, so that again their product is positive. If y = 0, then tan y= 0, hence the product sec y tan y is 0. Therefore, that product is never negative."

Let us then further note that $\tan(u) = \sqrt{1-\sec^2(u)}$ from the well known Pythagorean identity for $\tan(u)$. Thus, we can rewrite our fraction as $$\frac{1}{\sec(y)\sqrt{\sec^2(y)-1}}$$ We then back substitute $\sec^2(y) = x$ to get $$\frac{1}{x\sqrt{x^2-1}}$$
We then note that the term inside the square root will always be positive due to the squaring of $x$, but we still need to make sure the $x$ term out front always stays positive, so we rewrite this as $$\frac{1}{|x|\sqrt{x^2-1}}$$
(Note: the inside of the square root is the opposite sign that you got... that is the only mistake you made)

$\endgroup$

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.