Answer
$$ y'= \left (\cosh x \ln x+ \frac{\sinh x}{x}\right) x^{\sinh x}$$
Work Step by Step
Recall that $(\sinh x)'=\cosh x$
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\ln x)'=\dfrac{1}{x}$
Since $ y= x^{\sinh x}$, then taking the $\ln $ on both sides, we get
$$\ln y=\sinh x \ln x.$$
Now, the derivative, by using the product rule, is given by
$$ y'/y= \cosh x \ln x+ \frac{\sinh x}{x}\Longrightarrow y'= y(\cosh x \ln x+ \frac{\sinh x}{x}) $$
and hence $$ y'= \left (\cosh x \ln x+ \frac{\sinh x}{x}\right) x^{\sinh x}.$$
