Question

Find the first and second derivatives of the following function:

Y = sinh^-1Ɵ

Answer 1

\[\newcommand{sech}{\mathbin{\text{sech}}}y=\sinh^{-1}\theta~~\implies~~\sinh y=\theta\]
Differentiating both sides with respect to \(\theta\), you have
\[\cosh y\frac{dy}{d\theta}=1~~\implies~~\color{blue}{\frac{dy}{d\theta}=\sech(\sinh^{-1}\theta)}\]
This form isn't exactly pretty, so let's try to find a different one.
\[\sinh y=\frac{e^y-e^{-y}}{2}=\theta~~\implies~~\frac{e^{2y}-2\theta e^y-1}{2e^y}=0\]
The numerator is now quadratic in \(e^y\), so setting \(t=e^y\) gives us the following quadratic, which we can readily solve:
\[t^2-2\theta t-1=0~~\implies~~t=\frac{2\theta\pm\sqrt{4\theta^2+4}}{2}=\theta\pm\sqrt{\theta^2+1}\]
Back-substituting, we get
\[e^y=\theta+\sqrt{\theta^2+1}~~\iff~~\color{red}{y=\ln(\theta+\sqrt{\theta^2+1})}\equiv \sinh^{-1}\theta\]
We omit the negative root because \(e^y>0\). Taking the derivative of the red expression gives us a much neater form of the blue expression above.
\[y=\ln(\theta+\sqrt{\theta^2+1})~~\implies~~\frac{dy}{d\theta}=\frac{1+\dfrac{2\theta}{2\sqrt{\theta^2+1}}}{\theta+\sqrt{\theta^2+1}}=\frac{1+\dfrac{\theta}{\sqrt{\theta^2+1}}}{\theta+\sqrt{\theta^2+1}}\]
Some algebra cleans this right up:
\[\frac{dy}{d\theta}=\frac{\dfrac{\sqrt{\theta^2+1}+\theta}{\sqrt{\theta^2+1}}}{\theta+\sqrt{\theta^2+1}}=\frac{1}{\sqrt{\theta^2+1}}\]
and that's your first derivative. Nothing is stopping you from ending with the blue expression, but finding the second derivative using the red form of the inverse hyperbolic sine would likely be easier to grasp.