0

The derivatives of the inverse trig functions in almost all the textbooks or online documents are not presented in format A, but format B. Are there reasons for why we do this or is it just out of tradition?

Format A should be a much easier expression to help students link these new inverse trig formula to previous known trig ones and remember them easily.

(A) ($\sin^{-1}x$)' = $(\sqrt{1-x^2})^{-1}$ , ($\tan^{-1}x$)' = $({1+x^2})^{-1}$ , ($\sec^{-1}x$)' = $(|x|\sqrt{x^2-1})^{-1}$

(B) $(\sin^{-1}x)$' = $\frac{1}{\sqrt{1-x^2}}$ , $(\tan^{-1}x)$' = $\frac{1}{1+x^2}$ , $(\sec^{-1}x)$' = $\frac{1}{|x|\sqrt{x^2-1}}$

Format A is almost identical to that of derivatives of trig functions $\to$

($\sin \alpha$)'= $\sqrt{1-x^{2}}$, ($\tan \alpha$)'= $1+x^{2}$, ($\sec \alpha$)' = $|x|\sqrt{x^2-1}$

(given $\sin \alpha=x$, $\tan \alpha=x$, $\sec \alpha=x$)

Format B involves fraction expression $\frac{1}{X}$, I almost always need to mentally do some conversion to link it to the above formulas from which they are derived.

Any comments welcome!

V.G
  • 4,196
Star Bright
  • 2,338
  • 3
    Some $40$ years of teaching experience says that students tend to be more comfortable with the fractional form than with negative exponents. And in any case it’s really the expressions $\sqrt{1-x^2}$, $1+x^2$, and $|x|\sqrt{x^2-1}$, which are the same in both versions, that provide the real mnemonic link. And your notation is flawed: you mean $(\sin^{-1}x)'$, not $\sin^{-1}x'$ (and similarly for the rest). – Brian M. Scott Mar 08 '21 at 02:07
  • Brian, thanks for the background and notation correction! There is really a reason for this :-). In that senses, we should also use $arcsinx$ rather than $sin^-1$ x ? – Star Bright Mar 08 '21 at 02:38
  • I don’t mind using $\sin^{-1}x$: it has a long tradition, it’s consistent with the usual notation for inverse functions, and there are times, especially when teaching the material, when one wants to emphasize that it really is an inverse function, not some completely new creation out of thin air. But I’m also comfortable using $\arcsin x$. – Brian M. Scott Mar 08 '21 at 02:41
  • 1
    The problem/issue with $\sin^{-1}(x)$ is that it clashes with with the notation $\sin^n(x)$ to mean $(\sin(x))^n$ for any $n$ except $n=-1$ (well, for negative exponents it is a bit ambiguous too, but that’s really the problem). If we switched to $(\sin(x))^n$, there would be no conflicts/issues. – Arturo Magidin Mar 08 '21 at 02:55
  • 1
    I always use arcsin myself, to avoid confusion. – johnnyb Mar 08 '21 at 03:11

0 Answers0