So as the title states I'd like to find the derivative. I've used different methods but upon looking at the formula I noticed a difference between the author's approach and mine.




$\frac{1}{a*\sqrt {\sinh^2(y)+1}}=$

Until now I understand the reasoning, however this next step the author makes little sense to me:

$\frac{1}{\sqrt {a^2+x^2}}=$

What happens between these steps?

Many thanks whomever might help me!

  • 1
    $\begingroup$ What does $a$ have to do with $\arg\sinh x$? $\endgroup$
    – Bernard
    Jan 11, 2018 at 0:31
  • $\begingroup$ I forgot to add the denominator for the function. sorry! $\endgroup$ Jan 11, 2018 at 0:36
  • $\begingroup$ $x/a=\sinh(y)$, then distribute the $a$ into the square root. $\endgroup$
    – Ian
    Jan 11, 2018 at 0:40
  • $\begingroup$ That's it. Thank you @Ian $\endgroup$ Jan 11, 2018 at 0:43
  • $\begingroup$ By $\sinh^{-1}(x/a)$ you seem to mean the reciprocal of $\sinh(x/a)$, but the usual meaning of that notation is the arc-hyperbolic-sine which is the inverse function of hyperbolic sine. Which do you actually want? It appears the author wants the second meaning but you want the first. $\endgroup$ Jan 11, 2018 at 1:49

3 Answers 3


Let $y = \sinh^{-1} (x/a)$, then

$$ \begin{align} \sinh y &= \frac{x}{a} \\ \cosh y \ \frac{dy}{dx} &= \frac{1}{a} \\ \frac{dy}{dx} &= \frac{1}{a\sqrt{1+\sinh^2 y}} \\ &= \frac{1}{a\sqrt{1 + \dfrac{x^2}{a^2}}} \\ &= \frac{1}{\sqrt{a^2 + x^2}} \end{align} $$


That's pretty simple: $\;y=\arg\sinh \dfrac xa\iff \sinh y=\dfrac xa$, so $$\cosh^2y=1+\sinh^2y=1+\frac{x^2}{a^2}=\frac{x^2+a^2}{a^2},$$ and $$\frac{\mathrm d}{\mathrm dx}\bigl(\arg\sinh(x/a)\bigr)=\frac{1}{a\sqrt {\cosh y}}=\frac{1}{\not\mkern-2mu a\,\cfrac{\sqrt{x^2+a^2}}{\not\mkern-2mu a}}. $$


Since $$y= sinh^{-1}(x/a)$$ we get $$sinh^2(y)=(\frac {x}{a})^2$$

Therefore $$\frac{1}{a\sqrt {sinh^2(y)+1}}= \frac{1}{a\sqrt {(x/a)^2+1}}= \frac{1}{\sqrt{a^2 + x^2}} $$


Not the answer you're looking for? Browse other questions tagged .