Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 36

$y=x+1$

The tangent line at $x=0$ has slope $m=f^{\prime}(0)$ For $f^{\prime}(x)$, apply $\displaystyle \frac{d}{dx}[\sinh u]=(\cosh u)u^{\prime}$ and $\displaystyle \frac{d}{dx}[e^{u}]=e^{u}\cdot u^{\prime}$ $y=f(x)=e^{\sinh x}$ $f^{\prime}(x)=e^{\sinh x}\cdot\cosh x$ $m=f^{\prime}(0)=e^{0}(1)=1$ At $(0,1),$ the tangent line is: $y-y_{1}=m(x-x_{1})$ $y-1=1(x-0)$ $y=x+1$