## Calculus (3rd Edition)

$$y'=\frac{(1+\tanh t) (-csch^2 t)-\coth t( \, sech^2 t)}{(1+\tanh t)^2} .$$
Since $y=\frac{\coth t}{1+\tanh t}$, then the derivative $y'$, using the quotient rule $(u/v)'=\frac{vu'-uv'}{v^2}$ and the facts that $(\tanh t)'=sech^2t$ $(\coth t)'=-csch^2t$, is given by $$y'=\frac{(1+\tanh t) (-csch^2 t)-\coth t( \, sech^2 t)}{(1+\tanh t)^2} .$$