Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 33

$y=-2x+2$

The tangent line at $x=1$ has slope $m=f^{\prime}(1)$ For $f^{\prime}(x)$, apply the Chain Rule and $\displaystyle \frac{d}{dx}[\sinh u]=(\cosh u)u^{\prime}$ $f(x)=\sinh(1-x^{2}),$ $f^{\prime}(x)=\cosh(1-x^{2})(-2x)$ $m=f^{\prime}(1)=\cosh(0)(-2)=1(-2)=-2$ Tangent line at $(1,0)$: $y-y_{1}=m(x-x_{1})$ $y-0=-2(x-1)$ $y=-2x+2$