## Calculus (3rd Edition)

$$y'= \left (\cosh x \ln x+ \frac{\sinh x}{x}\right) x^{\sinh x}$$
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 $$y'= \left (\cosh x \ln x+ \frac{\sinh x}{x}\right) x^{\sinh x}.$$