Answer
$\DeclareMathOperator{\csch}{csch}$
$$g'(t)=\coth{\sqrt{t^2+1}}-\frac{t^2\csch^2{\sqrt{t^2+1}}}{\sqrt{t^2+1}}$$
Work Step by Step
$g'(t)=\frac{d}{dt}t\coth{\sqrt{t^2+1}}$
Using the product rule:
$g'(t)=1\times\coth{\sqrt{t^2+1}}+t\times\frac{d}{dt}\coth{\sqrt{t^2+1}}$
Using the chain rule:
$g'(t)=\coth{\sqrt{t^2+1}}+t(\frac{d\coth{\sqrt{t^2+1}}}{d\sqrt{t^2+1}}
\times \frac{d\sqrt{t^2+1}}{dt^2+1}
\times \frac{dt^2+1}{dt})$
$\DeclareMathOperator{\csch}{csch}$
$=\coth{\sqrt{t^2+1}}+t(-\csch^2{\sqrt{t^2+1}}
\times\frac{1}{2\sqrt{t^2+1}}
\times (2t)$
$=\coth{\sqrt{t^2+1}}-\frac{t^2\csch^2{\sqrt{t^2+1}}}{\sqrt{t^2+1}}$
