Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Then the derivative of the given function is $\frac{1}{\sqrt{1-(\frac{x+\sqrt{1-x^2}}{\sqrt 2}})^2}\cdot \frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Simplifying this gives the final answer $\frac{1}{\sqrt{1-x^2}}$. But the simplication process is quite lengthy and involves some bizarre calculations. Is there tricks/ways to solve these kinds of derivatives that do not involve too much calculations like above?

If you're confused about why your own method seemingly missed that the derivative is discontinuous, see here. The magic happens in line $4$. We have
$$\begin{split}
f'(x) 
&= \frac{1}{\sqrt{1-\left(\frac{x+\sqrt{1-x^2}}{\sqrt 2}\right)^2}}\cdot \frac{1}{\sqrt 2}\left(1-\frac{x}{\sqrt{1-x^2}}\right) \\
&= [\ldots] \\
&= \frac{1}{\sqrt{1-x^2}}\cdot \frac{\sqrt{1-x^2} - x}{\sqrt{\left(\sqrt{1-x^2} - x\right)^2}} \\
&= \frac{1}{\sqrt{1-x^2}}\cdot \frac{\sqrt{1-x^2} - x}{\left\lvert\sqrt{1-x^2} - x\right\rvert} \\
&= \frac{1}{\sqrt{1-x^2}}\cdot \operatorname{sign}\left(\sqrt{1-x^2} - x\right).
\end{split} $$
