Answer
$y=-2x+1$
Work Step by Step
The tangent line at $x=0$ has slope $m=f^{\prime}(0)$
For $f^{\prime}(x)$, apply
$[\sinh u]^{\prime}=(\cosh u)u^{\prime}$ and $[\cosh u]^{\prime}=(\sinh u)u^{\prime}$
$y=f(x)=(\cosh x-\sinh x)^{2}$
... chain rule for differentiation
$[u^{2}]^{\prime}=2u\cdot u^{\prime}$
$f^{\prime}(x)=2(\cosh x-\sinh x)\cdot(\sinh x-\cosh x)$
$m=f^{\prime}(0)=2(1)(-1)=-2$
At $(0,1),$ the tangent line is:
$y-y_{1}=m(x-x_{1})$
$y-1=-2(x-0)$
$y=-2x+1$
