Answer
$\DeclareMathOperator{\csch}{csch}$$\coth{\sqrt{t^2+1}}-\dfrac{t^2\csch^2{\sqrt{t^2+1}}}{\sqrt{t^2+1}}$
Work Step by Step
Given: $g(t)=t\coth{\sqrt{t^2+1}}$ Now, $g'(t)=\dfrac{d}{dt}(t\coth{\sqrt{t^2+1}})$ Apply product rule, we have $g'(t)=(1)\coth{\sqrt{t^2+1}}+t\times\dfrac{d}{dt}[\coth{\sqrt{t^2+1}}]$ Now, apply chain rule. $g'(t)=\coth{\sqrt{t^2+1}}+t(\frac{d\coth{\sqrt{t^2+1}}}{d\sqrt{t^2+1}}(\dfrac{d\sqrt{t^2+1}}{dt^2+1})(\dfrac{dt^2+1}{dt}))$ This implies, $\DeclareMathOperator{\csch}{csch}$$g'(t)=\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}}-\dfrac{t^2\csch^2{\sqrt{t^2+1}}}{\sqrt{t^2+1}}$
