Answer
$$ y'=\cosh x\tanh x+\sinh x\, sech^2 x .$$
Work Step by Step
Since $ y=\sinh x\tanh x $, then the derivative $ y'$, using the product rule $(uv)'=uv'+u'v $ and the facts that $(\sinh x)'=\cosh x $ and $(\tanh x)'=sech^2x $, is given by $$ y'=\cosh x\tanh x+\sinh x\, sech^2 x .$$
