Simplify $f(x)=\sec (\tan^{-1} (\sin (\tan^{-1} x)))$ for $x\in\mathbb{R}$ and find its inflection points 
Simplify $f(x)=\sec (\tan^{-1} (\sin (\tan^{-1} x)))$ for $x\in\mathbb{R}$ and find its inflection points, local extrema, x-intercepts, y-intercept, asymptotes, etc.

So the $y$-intercept is obviously $1$. 
I'm not exactly sure if my simplification below is correct, but if it is, everything else that's required should be trivial for me (finding the local extrema and inflections points is very easy, though I'd like to know if there are any shortcuts other than evaluating the derivatives and relevant limits).
So I know by drawing a simple triangle that $\sin(\tan^{-1}x)) = \dfrac{x}{\sqrt{1+x^2}}$ and 
$\sec(\tan^{-1}x)=\sqrt{1+x^2}.$ But then, doesn't that mean that $\sec(\tan^{-1}(\sin(\tan^{-1}x)))=\sqrt{2-\dfrac{1}{1+x^2}}$?
 A: Note, you have $y=\sin(\tan^{-1}x)) = \dfrac{x}{\sqrt{1+x^2}}$. Then,
$$\sec(\tan^{-1}(\sin(\tan^{-1}x)))=\sec(\tan^{-1}y)=\sqrt{1+y^2}$$
$$=\sqrt{1+\dfrac{x^2}{1+x^2}}=\sqrt{\dfrac{1+2x^2}{1+x^2}}$$
A: Note that $$f'(x)=\frac{x}{\left(x^2+1\right)^2 \sqrt{\frac{2
   x^2+1}{x^2+1}}}$$ and $$f''(x)=-\frac{6 x^4+2 x^2-1}{\left(x^2+1\right)^4
   \left(\frac{2 x^2+1}{x^2+1}\right)^{3/2}}$$
Solve the equation $$f''(x)=0$$ to get inflection points.
A: You are right on computing $f(x)$. Now $f$ obviously is even function and is strictly monotone before and after origin. In fact it is strictly decreasing before it, has a minimum at origin $(0,1)$ and then strictely increasing after it. In both sides tends to $\sqrt 2$, so $y=\sqrt 2$ is horizontal asymptote in both side that $f$ increasingly tends to it. But for inflection points I have nothing to say without derivating, except we have two of them that are in $(\pm x_0,y_0)$. By what @Dr.SonnhardGraubner compute, we find $x_0 = \sqrt\frac{\sqrt 7 -1}{6}$, and $y_0 = f(x_0)$.
