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