Derivative of following functions $f(x) = \cos^{-1}\left(1-2x^2\right)$ Derivative of following functions 
(A) $f(x) = \cos^{-1}\left(1-2x^2\right)$
(B) $f(x) = \sin^{-1}(3x-4x^3)$
$\bf{My\; Try}::$ Given $f(x) = \cos^{-1}\left(1-2x^2\right)$ and function $f(x)$ is defined when $\left|1-2x^2\right|\leq 1\Rightarrow -1 \leq x\leq 1$
Now put $x=\sin \theta$, where $\displaystyle -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\Rightarrow -\pi\leq 2\theta \leq \pi$ 
we get $\displaystyle f(x) = \cos^{-1}(1-2\sin^2 \theta) = \cos^{-1}(\cos 2 \theta)$
$ = -2\theta \;\;\;, -\pi \leq 2\theta \leq 0$
$=+2\theta \;\;\;, 0 \leq 2\theta \leq \pi$ 
$ f(x) = -2\sin^{-1}(x) \;\;\;, -1 \leq x \leq 0$
$\;\;\; \;\;\;\; =+2\sin^{-1}(x) \;\;\;, 0 \leq x \leq 1$
So $\displaystyle  f^{'}(x) = -\frac{2}{\sqrt{1-x^2}} \;\;\;, -1 \leq x \leq 0$
$\displaystyle \;\;\; \;\;\;\; =+\frac{2}{\sqrt{1-x^2}} \;\;\;, 0 \leq x \leq 1$
So function $f(x) = \cos^{-1}(1-2x^2)$ is not Differentiable at $x = 0$
Is my process is Right , If not Then How can we solve it
Help me
Thanks

Your process seems to be fine.
Putting $x=\cos\theta,0\le \theta\le \pi$ as the principal value of inverse cosine function
I also reached at $$\cos^{-1}(1-2x^2)=\begin{cases} 2\cos^{-1}x-\pi &\mbox{if } \frac\pi2\le \theta\le \pi\implies0\ge x\ge -1 \\
\pi- 2\cos^{-1}x & \mbox{if } 0\le \theta\le \frac\pi2\implies1\ge x\ge0 \end{cases} $$
Observe that $$f'(x)=\frac{d\{\cos^{-1}(1-2x^2)\}}{dx}=\frac{d\{\cos^{-1}(1-2x^2)\}}{d(1-2x^2)}\cdot\frac{d(1-2x^2)}{dx}$$
$$=\frac{-1}{\sqrt{1-(1-2x^2)^2}}\cdot(-4x)=\frac{2x}{\sqrt{x^2-x^4}}$$
$$=\frac{2x}{|x|\sqrt{1-x^2}}\begin{cases} =\frac2{\sqrt{1-x^2}}  &\mbox{if } |x|=x\iff x>0 \\ =-\frac2{\sqrt{1-x^2}}  &\mbox{if } |x|=-x\iff x<0  &\\ \text{   is clearly undefined  } &\mbox{if } x=0\end{cases} $$ 
