# Hyperbolic differentiation of $\sinh^{-1}(x/a)$ [closed]

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.

so

$\frac{d}{dx}\sinh^{-1}(x/a)=$

$\frac{1}{a*\cosh(y)}=$

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

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!

• What does $a$ have to do with $\arg\sinh x$? Jan 11, 2018 at 0:31
• I forgot to add the denominator for the function. sorry! Jan 11, 2018 at 0:36
• $x/a=\sinh(y)$, then distribute the $a$ into the square root.
– Ian
Jan 11, 2018 at 0:40
• That's it. Thank you @Ian Jan 11, 2018 at 0:43
• 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. Jan 11, 2018 at 1:49

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