Answer
$$\frac{d}{d t} \sinh ^{-1} t=\frac{1}{\sqrt{t^{2}+1}} $$
Work Step by Step
Let
\begin{align*}
x&=\sinh ^{-1} t \Rightarrow t=\sinh x
\end{align*}
then
\begin{align*}
\frac{d t}{d x}&=\cosh x\\
\frac{d x}{d t}&=\frac{1}{\cosh x}\\
&=\frac{1}{\sqrt{\sinh^2x+1}}\\
&=\frac{1}{\sqrt{t^2 +1}}
\end{align*}
Hence, verified.
