Domain of $y'$ if $y=e^{\sec^2x}+3\cos^{-1}x$
It is said that the derivative of the given function is valid in $[-1,1]-\{0\}$.
$$ y'=e^{\sec^2x}\cdot2\sec x\cdot\sec x\tan x-\frac{3}{\sqrt{1-x^2}}=2\sec^2x\cdot\tan x\cdot e^{\sec^2x}-\frac{3}{\sqrt{1-x^2}} $$ $$ 1-x^2>0\implies x^2<1\implies|x|<1\implies -1<x<1\implies x\in(-1,1) $$ Both $\sec x$ and $\tan x$ are defined for $x=0$. Then how can it be $[-1,1]-\{0\}$ ?
ok the first derivative is given by $$f'(x)=2\, \left( \sec \left( x \right) \right) ^{2}\tan \left( x \right) { {\rm e}^{ \left( \sec \left( x \right) \right) ^{2}}}-3\,{\frac {1}{ \sqrt {-{x}^{2}+1}}} $$ from the term with the square root we get $$(1-x)(1+x)>0$$ and $$\cos(x)\ne 0$$