Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 30

$y^{\prime}=x\sinh x$

Apply the product rule andTh. 5.18:$\qquad$ $ \displaystyle \frac{d}{dx}[\sinh u]=(\cosh u)u^{\prime}$ $\displaystyle \frac{d}{dx}[\cosh u]=(\sinh u)u^{\prime}$ $y=x\cosh x-\sinh x$ $y^{\prime}=[x(\cosh x)^{\prime}+(x)^{\prime}\cosh x]-(\sinh x)^{\prime}$ $y^{\prime}=[x\sinh x+\cosh x]-\cosh x$ $y^{\prime}=x\sinh x$